5 grams of a sample of magnesium carbonate on treatment with excess of

1.	5 grams of a sample of magnesium carbonate on treatment with excess of dilute hydrochloric acid give 1.12 litre of Co2 at STP the percentage of magnesium carbonate in the mixture is
Answer» Answer: 83.3 % Explanation: According to avogadro's law, 1 mole of EVERY substance OCCUPIES 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles. To calculate the moles, we USE the equation: $\text{Number of moles of magnesium}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{5g}{84g/mol}=0.06moles$ $MgCO_3\rightarrow MgO+CO_2$ 1 mole of magnesium carbonate produce= 22.4 L of carbon dioxide at STP Thus 0.06 moles of magnesium produce= $\frac{22.4}{1}\times 0.06=1.344L$ Thus percentage PURITY= $\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{1.12}{1.344}\times 100=83.3\%$ $MgCO_3$ Therefore, the percentage purity of is 83.3%.

5 grams of a sample of magnesium carbonate on treatment with excess of dilute hydrochloric acid give 1.12 litre of Co2 at STP the percentage of magnesium carbonate in the mixture is

Answer»

Answer: 83.3 %

Explanation:

According to avogadro's law, 1 mole of EVERY substance OCCUPIES 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we USE the equation:

$\text{Number of moles of magnesium}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{5g}{84g/mol}=0.06moles$

$MgCO_3\rightarrow MgO+CO_2$

1 mole of magnesium carbonate produce= 22.4 L of carbon dioxide at STP

Thus 0.06 moles of magnesium produce= $\frac{22.4}{1}\times 0.06=1.344L$

Thus percentage PURITY= $\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{1.12}{1.344}\times 100=83.3\%$

$MgCO_3$ Therefore, the percentage purity of is 83.3%.