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Purity of calcium carbonate in limestone
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+ Stoichiometry
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Muhammad Shakaib
Purity of calcium carbonate in limestone
We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.
Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$
We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.
Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$
We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.
Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$
...implying the purity is 75%.
We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.
Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$
...implying the purity is 75%.
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