Question

112 cm of $H_{2}S$ (g) is mixed with $120c{m}^3$ (g) at STP to produce HCl(g) and sulphur (s).

Write a balanced Equation for this reaction and calculate (i) the volume of gaseous product formed (ii) composition of the resulting mixture.

Answer 1

$H_{2}S+C{l}_2⟶2HCl+S$

when 22.4L of $H_{2}S$ reacts with 22.4L of $C{l}_2$ it produce 32 g sulphur

$1c{m}^3=0.001L$

$112c{m}^3$ of $H_{2}S=0.112L$ of $H_{2}S$

$120c{m}^{3}ofC{l}_2=0.120LofC{l}_2$

$112c{m}^3$ of $H_{2}S$ will react with only 0.112L of $C{l}_2$

22.4L of $H_{2}S$ produce 32 g of S

∴ 0.112 L of $H_{2}S$ will produce $\frac{32}{22.4}\times 0.112$ = 0.16g of Sulfur

Unused chlorine = $(120-112)c{m}^3$ = 8$c{m}^3$

22.4 l of $H_{2}S$ produce 2$\times$22.4L HCL

$112c{m}^3$ of $H_{2}S$ will produce 2$\times$ 112 $c{m}^3$ of HCl

resultimg mixture = 8$c{m}^3$ of chlorine, 0.16 g of sulphur and 224$c{m}^3$ of HCL