Question

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 1

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

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

Answer 2

To caluculate the no of moles

We use the formula:Given Mass / Molar Mass

Then no of moles of magnesium is 5/84=0.06 moles . We know that the volume of any gas occupies 22.4 litres of STP then

:0.06×22.4= 1.344 litres of Volume

We now that : Purity of Substance :Theoratical volume /actual volume

:then = 1.12/1.344

:0.83

Then the percentage of magnesium carbonate in the mixture is : Purity of Substance×100

: 0.83×100

: 83%

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