Some Basic Concepts of Chemistry - Result Question 4
####7. When 22.4 litres of $H_2(g)$ is mixed with 11.2 litres of $Cl_2(g)$, each at S.T.P., the moles of $HCl(g)$ formed is equal to :
[2014]
(a) 1 mole of $HCl(g)$
(b) 2 moles of $HCl(g)$
(c) 0.5 moles of $HCl(g)$
(d) 1.5 moles of $HCl$ (g)
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Solution:
- (a) $H_2$ (g) $+Cl_2$ (g) $\longrightarrow 2 HCl(g)$
(4) Moles of water $=\frac{0.00224}{22.4}=10^{-4}$
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