1.

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

Answer»

1 mol of HCl (g)
2 mol of HCl (g)
0.5 mol ofHCl (g)
1.5 mol of HCl (g)

SOLUTION :22.4 L of `H_(2)` = 1 mol of `H_(2)`
11.2 L of `Cl_(2)=(11.2)/(22.4)"mol = 0.5 mol of "Cl_(2)`
`UNDERSET("1 mol")(H_(2)(g))+underset("1 mol")(Cl_(2)(g))rarrunderset("2 mol")(2HCl)(g)`
Thus, `Cl_(2)` will be the limiting reagent and HCl formed by reaction of 0.5 mol of `Cl_(2)=1` mol


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