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What kind of reaction is this? H2SO4 + CuO-> CuSO4 + H2O?
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Linda C. Ragin
What kind of reaction is this? H2SO4 + CuO-> CuSO4 + H2O?
To classify a reaction, look at the numbers and types of reactants and products.
In this reaction you have two reactants forming two products with no change in oxidation states. That is a sign that you have a “ double displacment” or “ metathesis” reaction.
The two H atoms, both with a + 1 oxidation state, have changed places with the Cu ^+2.
The most familiar double displacement reactions occur in aqueous solution, but that is not a requirement. This type if reaction is sometimes used to dissolve an insoluble solid.
To classify a reaction, look at the numbers and types of reactants and products.
In this reaction you have two reactants forming two products with no change in oxidation states. That is a sign that you have a “ double displacment” or “ metathesis” reaction.
The two H atoms, both with a + 1 oxidation state, have changed places with the Cu ^+2.
The most familiar double displacement reactions occur in aqueous solution, but that is not a requirement. This type if reaction is sometimes used to dissolve an insoluble solid.
However, if you combine H[math]_{2}[/math]O with CuSO[math]_{4}[/math] in a molar ratio of greater than 2:1 you will form some combination of aqueous CuSO[math]_{4}[/math] and solid CuSO[math]_{4}[/math].2H[math]_{2}[/math]O. At ratios above the saturation point, only aqueous CuSO[math]_{4}[/math] (Cu[math]^{+2}[/math] & SO[math]_{4}^{-2}[/math]) will exist.
However, if you combine H[math]_{2}[/math]O with CuSO[math]_{4}[/math] in a molar ratio of greater than 2:1 you will form some combination of aqueous CuSO[math]_{4}[/math] and solid CuSO[math]_{4}[/math].2H[math]_{2}[/math]O. At ratios above the saturation point, only aqueous CuSO[math]_{4}[/math] (Cu[math]^{+2}[/math] & SO[math]_{4}^{-2}[/math]) will exist.
It is a double displacement/double decomposition reaction, because both the cations and anions in both reactants have exchanged.
It is also a neutralization reaction. Because acid [math]H_{2}SO_{4}[/math] reacts with base [math]CuO[/math] to produce a salt [math]CuSO_{4}[/math] and water.
It is a double displacement/double decomposition reaction, because both the cations and anions in both reactants have exchanged.
It is also a neutralization reaction. Because acid [math]H_{2}SO_{4}[/math] reacts with base [math]CuO[/math] to produce a salt [math]CuSO_{4}[/math] and water.
Jim Engel is correct. You will get an acidic solution of copper sulfate pentahydrate, CuSO4•5H2O. Certainly, there is no chemical reaction between H2SO4 and CuSO4•5H2O (we are using the pentahydrate form, since this is a solution – presumably in water).
Jim Engel is correct. You will get an acidic solution of copper sulfate pentahydrate, CuSO4•5H2O. Certainly, there is no chemical reaction between H2SO4 and CuSO4•5H2O (we are using the pentahydrate form, since this is a solution – presumably in water).
It is called a “double replacement” (double displacement in UK) reaction. Since CuO is basic in nature, it can also be considered acid/base neutralization.
It is called a “double replacement” (double displacement in UK) reaction. Since CuO is basic in nature, it can also be considered acid/base neutralization.
In a highly exothermic process, sulfuric acid (H2SO4) interacts strongly with water. When water is added to concentrated sulfuric acid, it can boil and spit, resulting in a painful acid burn.
In a highly exothermic process, sulfuric acid (H2SO4) interacts strongly with water. When water is added to concentrated sulfuric acid, it can boil and spit, resulting in a painful acid burn.
Yes, BUT you will not see it if you stick a piece of copper into sulfuric acid because the equilibrium is very far to the left. What you will see is the copper oxide being removed from the copper because it is soluble in acid. Jewelers use NaHSO4 solutions, called ‘pickle’ to remove fire scale (copper oxide) from jewelry after soldering.
Yes, BUT you will not see it if you stick a piece of copper into sulfuric acid because the equilibrium is very far to the left. What you will see is the copper oxide being removed from the copper because it is soluble in acid. Jewelers use NaHSO4 solutions, called ‘pickle’ to remove fire scale (copper oxide) from jewelry after soldering.
To classify a reaction, look at the numbers and types of reactants and products.
In this reaction you have two reactants forming two products with no change in oxidation states. That is a sign that you have a “ double displacment” or “ metathesis” reaction.
The two H atoms, both with a + 1 oxidation state, have changed places with the Cu ^+2.
The most familiar double displacement reactions occur in aqueous solution, but that is not a requirement. This type if reaction is sometimes used to dissolve an insoluble solid.
To classify a reaction, look at the numbers and types of reactants and products.
In this reaction you have two reactants forming two products with no change in oxidation states. That is a sign that you have a “ double displacment” or “ metathesis” reaction.
The two H atoms, both with a + 1 oxidation state, have changed places with the Cu ^+2.
The most familiar double displacement reactions occur in aqueous solution, but that is not a requirement. This type if reaction is sometimes used to dissolve an insoluble solid.
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Zn + H2SO4 ———-> ZnSO4 + H2 is a Single displacement reaction
Thank you, hope you got your answer :-)
Zn + H2SO4 ———-> ZnSO4 + H2 is a Single displacement reaction
Thank you, hope you got your answer :-)
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It really depends on the ratios of CuSO[math]_{4}[/math] and H[math]_{2}[/math]O.
If you combine H[math]_{2}[/math]O and CuSO[math]_{4}[/math] in a molar ratio of 2:1, you will form crystalline copper sulphate dihydrate;-
CuSO[math]_{4}[/math] + 2 H[math]_{2}[/math]O → CuSO[math]_{4}[/math].2H[math]_{2}[/math]O
However, if you combine H[math]_{2}[/math]O with CuSO[math]_{4}[/math] in a molar ratio of greater than 2:1 you will form some combination of aqueous CuSO[math]_{4}[/math] and solid CuSO[math]_{4}[/math].2H[math]_{2}[/math]O. At ratios above the saturation point, only aqueous CuSO[math]_{4}[/math] (Cu[math]^{+2}[/math] & SO[math]_{4}^{-2}[/math]) will exist.
It really depends on the ratios of CuSO[math]_{4}[/math] and H[math]_{2}[/math]O.
If you combine H[math]_{2}[/math]O and CuSO[math]_{4}[/math] in a molar ratio of 2:1, you will form crystalline copper sulphate dihydrate;-
CuSO[math]_{4}[/math] + 2 H[math]_{2}[/math]O → CuSO[math]_{4}[/math].2H[math]_{2}[/math]O
However, if you combine H[math]_{2}[/math]O with CuSO[math]_{4}[/math] in a molar ratio of greater than 2:1 you will form some combination of aqueous CuSO[math]_{4}[/math] and solid CuSO[math]_{4}[/math].2H[math]_{2}[/math]O. At ratios above the saturation point, only aqueous CuSO[math]_{4}[/math] (Cu[math]^{+2}[/math] & SO[math]_{4}^{-2}[/math]) will exist.
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It is an addition reaction.
It is a typical one for when the anhydride of an acid reacts with water to produce the acid.
Other reactions of this exact type are, for example:
It is an addition reaction.
It is a typical one for when the anhydride of an acid reacts with water to produce the acid.
Other reactions of this exact type are, for example:
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[math]H_{2}SO_{4}+CuO=CuSO_{4}+H_{2}O[/math]
It is a double displacement/double decomposition reaction, because both the cations and anions in both reactants have exchanged.
It is also a neutralization reaction. Because acid [math]H_{2}SO_{4}[/math] reacts with base [math]CuO[/math] to produce a salt [math]CuSO_{4}[/math] and water.
[math]H_{2}SO_{4}+CuO=CuSO_{4}+H_{2}O[/math]
It is a double displacement/double decomposition reaction, because both the cations and anions in both reactants have exchanged.
It is also a neutralization reaction. Because acid [math]H_{2}SO_{4}[/math] reacts with base [math]CuO[/math] to produce a salt [math]CuSO_{4}[/math] and water.
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Jim Engel is correct. You will get an acidic solution of copper sulfate pentahydrate, CuSO4•5H2O. Certainly, there is no chemical reaction between H2SO4 and CuSO4•5H2O (we are using the pentahydrate form, since this is a solution – presumably in water).
Jim Engel is correct. You will get an acidic solution of copper sulfate pentahydrate, CuSO4•5H2O. Certainly, there is no chemical reaction between H2SO4 and CuSO4•5H2O (we are using the pentahydrate form, since this is a solution – presumably in water).
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It is called a “double replacement” (double displacement in UK) reaction. Since CuO is basic in nature, it can also be considered acid/base neutralization.
It is called a “double replacement” (double displacement in UK) reaction. Since CuO is basic in nature, it can also be considered acid/base neutralization.
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CuO + H2SO4 = CuSO4 + H2O(l)
Change in Free Energy: ΔG(20C) = -79.9kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -85.9kJ (negative, so the reaction is exothermic)
This is a double displacement, exothermic reaction.
CuO + H2SO4 = CuSO4 + H2O(l)
Change in Free Energy: ΔG(20C) = -79.9kJ (negative, so the reaction runs)
Change in Enthalpy: ΔH(20C) = -85.9kJ (negative, so the reaction is exothermic)
This is a double displacement, exothermic reaction.
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In a highly exothermic process, sulfuric acid (H2SO4) interacts strongly with water. When water is added to concentrated sulfuric acid, it can boil and spit, resulting in a painful acid burn.
In a highly exothermic process, sulfuric acid (H2SO4) interacts strongly with water. When water is added to concentrated sulfuric acid, it can boil and spit, resulting in a painful acid burn.
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Yes, BUT you will not see it if you stick a piece of copper into sulfuric acid because the equilibrium is very far to the left. What you will see is the copper oxide being removed from the copper because it is soluble in acid. Jewelers use NaHSO4 solutions, called ‘pickle’ to remove fire scale (copper oxide) from jewelry after soldering.
Yes, BUT you will not see it if you stick a piece of copper into sulfuric acid because the equilibrium is very far to the left. What you will see is the copper oxide being removed from the copper because it is soluble in acid. Jewelers use NaHSO4 solutions, called ‘pickle’ to remove fire scale (copper oxide) from jewelry after soldering.
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Your question is incomplete.
The limiting reagent in any reaction is the one that has the fewest molar equivalents for that reaction.
Your question is incomplete.
The limiting reagent in any reaction is the one that has the fewest molar equivalents for that reaction.
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