Home >
Community >
What is the theoretical yield of rhodium(III) hydroxide from the reaction of 0.540 g of rhodium (III) sulfate with 0.209 g of sodium hydroxide? Equation: RH2(SO4) 3 + 6NaOH -> 2Rh(OH) 3 + 3Na2SO4.
Upvote
15
Downvote
+ Chemical reactions
+ Hydroxides
+ Sodium hydroxide
+ Chemistry
+ Stoichiometry
Posted by
Mike Eisler
What is the theoretical yield of rhodium(III) hydroxide from the reaction of 0.540 g of rhodium (III) sulfate with 0.209 g of sodium hydroxide? Equation: RH2(SO4) 3 + 6NaOH -> 2Rh(OH) 3 + 3Na2SO4.
In this question you are given an experimental mass of both of the reactants. We have to determine which reactant is limiting (will be all used up) because this reactant will determine how much product we can produce.
First start by determining the number of moles of each reactant that is being used in this experiment:
To determine which reactant is limiting, we must compare the experimental mole ratio to the ideal ratio defined by the balanced chemical equation:
Ideal Ratio for NaOH/Rh2(SO4)3 = 6 mol/1mol
Exp. Ratio for NaOH/Rh2(SO4)3 = 0.005225 mol/0.001093 mol = 4.78 mol/1 mol
This tells us that there is not enough NaOH to consume all of the Rh2(SO4)3. The NaOH is the limiting reactant (the reaction will stop when the NaOH is all used up).
The number of moles of Rh(OH)3 that should be produced can be predicted by assuming that all of the NaOH is converted to product and using the mole ratio from the balanced chemical equation:
n(Rh(OH)3) = 0.005225 mol NaOH x (2 mol Rh(OH)3/6 mol NaOH) = 0.001742 mol Rh(OH)3 produced
We can now convert moles of Rh(OH)3 to mass using its molar mass:
m(Rh(OH)3) = n • M = 0.001742 mol (153.98 g/mol) = 0.268 g Rh(OH)3
In this question you are given an experimental mass of both of the reactants. We have to determine which reactant is limiting (will be all used up) because this reactant will determine how much product we can produce.
First start by determining the number of moles of each reactant that is being used in this experiment:
To determine which reactant is limiting, we must compare the experimental mole ratio to the ideal ratio defined by the balanced chemical equation:
Ideal Ratio for NaOH/Rh2(SO4)3 = 6 mol/1mol
Exp. Ratio for NaOH/Rh2(SO4)3 = 0.005225 mol/0.001093 mol = 4.78 mol/1 mol
This tells us that there is not enough NaOH to consume all of the Rh2(SO4)3. The NaOH is the limiting reactant (the reaction will stop when the NaOH is all used up).
The number of moles of Rh(OH)3 that should be produced can be predicted by assuming that all of the NaOH is converted to product and using the mole ratio from the balanced chemical equation:
n(Rh(OH)3) = 0.005225 mol NaOH x (2 mol Rh(OH)3/6 mol NaOH) = 0.001742 mol Rh(OH)3 produced
We can now convert moles of Rh(OH)3 to mass using its molar mass:
m(Rh(OH)3) = n • M = 0.001742 mol (153.98 g/mol) = 0.268 g Rh(OH)3
In this question you are given an experimental mass of both of the reactants. We have to determine which reactant is limiting (will be all used up) because this reactant will determine how much product we can produce.
First start by determining the number of moles of each reactant that is being used in this experiment:
n(Rh2(SO4)3) = m/M = 0.540 g/493.999 g/mol = 0.001093 mol Rh2(SO4)3
n(NaOH) = m/M = 0.209 g/39.997 g/mol = 0.005225 mol NaOH
To determine which reactant is limiting, we must compare the experimental mole ratio to the ideal ratio defined by the balanced chemical equation:
Ideal Ratio for NaOH/Rh2(SO4)3 = 6 mol/1mol
Exp. Ratio for NaOH/Rh2(SO4)3 = 0.005225 mol/0.001093 mol = 4.78 mol/1 mol
This tells us that there is not enough NaOH to consume all of the Rh2(SO4)3. The NaOH is the limiting reactant (the reaction will stop when the NaOH is all used up).
The number of moles of Rh(OH)3 that should be produced can be predicted by assuming that all of the NaOH is converted to product and using the mole ratio from the balanced chemical equation:
n(Rh(OH)3) = 0.005225 mol NaOH x (2 mol Rh(OH)3/6 mol NaOH) = 0.001742 mol Rh(OH)3 produced
We can now convert moles of Rh(OH)3 to mass using its molar mass:
m(Rh(OH)3) = n • M = 0.001742 mol (153.98 g/mol) = 0.268 g Rh(OH)3
In this question you are given an experimental mass of both of the reactants. We have to determine which reactant is limiting (will be all used up) because this reactant will determine how much product we can produce.
First start by determining the number of moles of each reactant that is being used in this experiment:
n(Rh2(SO4)3) = m/M = 0.540 g/493.999 g/mol = 0.001093 mol Rh2(SO4)3
n(NaOH) = m/M = 0.209 g/39.997 g/mol = 0.005225 mol NaOH
To determine which reactant is limiting, we must compare the experimental mole ratio to the ideal ratio defined by the balanced chemical equation:
Ideal Ratio for NaOH/Rh2(SO4)3 = 6 mol/1mol
Exp. Ratio for NaOH/Rh2(SO4)3 = 0.005225 mol/0.001093 mol = 4.78 mol/1 mol
This tells us that there is not enough NaOH to consume all of the Rh2(SO4)3. The NaOH is the limiting reactant (the reaction will stop when the NaOH is all used up).
The number of moles of Rh(OH)3 that should be produced can be predicted by assuming that all of the NaOH is converted to product and using the mole ratio from the balanced chemical equation:
n(Rh(OH)3) = 0.005225 mol NaOH x (2 mol Rh(OH)3/6 mol NaOH) = 0.001742 mol Rh(OH)3 produced
We can now convert moles of Rh(OH)3 to mass using its molar mass:
m(Rh(OH)3) = n • M = 0.001742 mol (153.98 g/mol) = 0.268 g Rh(OH)3
More
VOTE
molRH2(SO4) 3=0.54gr/(206+3*32+12*16gr.mol^-1)=0.0011 excess reactant
molNaOH=0.209gr/40gr.mol^-1/6=8.7*10^-4 limiting reactant
grtheoretical yield of rhodium(III) hydroxide=8.7*10^-4 *2*(103+48+3gr.mol^-1)=0.2682gr
molRH2(SO4) 3=0.54gr/(206+3*32+12*16gr.mol^-1)=0.0011 excess reactant
molNaOH=0.209gr/40gr.mol^-1/6=8.7*10^-4 limiting reactant
grtheoretical yield of rhodium(III) hydroxide=8.7*10^-4 *2*(103+48+3gr.mol^-1)=0.2682gr
More
VOTE