Question

$70$ gram of a sample of magnesite on treatment with excess of HCl gave $11.2$ litre of $C{O}_2$ at STP. The percentage purity of the sample is?

• A. 80
• B. 70
• C. 60
• D. 50

Answer 1

Answer: (C)

The explanation for the correct option:

Option(C) 60

• It is given that mass of magnesite$\left(MgC{O}_3\right)=70g$
• The volume of $C{O}_2==11.2litre$
• The reaction is given as:

  $MgC{O}_3+2HCl⟶MgC{l}_2+C{O}_2+H_{2}O$

  $1mole1mole1mole1mole1mole$

  $84g46g95g44g18g$
  

• By Avogadro's law, 1 mole of every substance occupies 22.4 liters, then ​the number of moles of $C{O}_2$ occupying 11.2 liters is $\frac{11.2}{22.4}=0.5moles$

• The equation for the number of moles is given by $n=\frac{givenmassm}{molarmassM}$ then, $0.5=\frac{m}{44}\Rightarrow m=22g$

• All the reactants and products have same number of moles, hence, magnesite will also present as 0.5 moles

• Hence, the mass of magnesite $=0.5\times 84=42gofMgC{O}_3$

• Hence, 70 g of the sample contains 42 g of magnesite.

• Hence, the % of purity$=\frac{experimentalyield}{Theoroticalyield}\times 100=\frac{42}{70}\times 100=60\%$

Hence, Option (C) 60% is correct option