Use Mn = 55, O = 16 and molar volume $22.4 L/mol$ at STP. $MnO X_{2} + 4 HCl MnCl X_{2} + Cl X_{2} + 2 H X_{2} O$ . A mixture of $17.4 g$ of $MnO X_{2}$ and $500 mL$ of $1.2 M$ $HCl$ is heated. Find the volume of $Cl X_{2}$ evolved at STP in litres (two decimals), assuming complete reaction of the limiting reagent.

Numerical answer type, answer in L, tolerance 0.02.

Answer: 3.36 L

n (MnO X_{2}) = 17.4/87 = 0.20

mol;

n (HCl) = 0.5 \times 1.2 = 0.60

mol, which can react with only

0.60/4 = 0.150

mol

MnO X_{2}

. So HCl is limiting, giving

0.150

mol

Cl X_{2}

. Volume

= 0.150 \times 22.4 = 3.36 L