13 g of zinc is treated with 7.3 g of HCl according to Zn + 2HCl -> ZnCl2 + H2 (Zn = 65, H = 1, Cl = 35.5; 22.4 L mol^-1 at STP). The volume of H2 liberated at STP is:

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$13 g$ of zinc is treated with $7.3 g$ of HCl according to $Zn + 2 HCl ZnCl X_{2} + H X_{2}$ (Zn = 65, H = 1, Cl = 35.5; $22.4 L mo l^{- 1}$ at STP). The volume of $H X_{2}$ liberated at STP is:

n (Zn) = 13/65 = 0.20

;

n (HCl) = 7.3/36.5 = 0.20

. Zn needs

0.40

mol HCl, so HCl is limiting and gives

0.20/2 = 0.10

mol

H X_{2}

V = 0.10 \times 22.4 = 2.24 L

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