When 22.4 litres of is mixed with 11.2 litres of , each at STP, the moles of formed is equal to [2014]

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Q.
When 22.4 litres of $H_{2 (g)}$ is mixed with 11.2 litres of ${Cl}_{2 (g)}$ , each at STP, the moles of ${HCl}_{(g)}$ formed is equal to [2014]

1 1 mol of ${HCl}_{(g)}$

2 2 mol of ${HCl}_{(g)}$

3 0.5 mol of ${HCl}_{(g)}$

4 1.5 mol of ${HCl}_{(g)}$

Ans.
(1)

1 mole $\equiv$ 22.4 litres at STP.

$n_{H_{2}} = \frac{22.4}{22.4} = 1 mol; n_{{Cl}_{2}} = \frac{11.2}{22.4} = 0.5 mol$

Reaction is as,

Here, $C l_{2}$ is the limiting reagent. So, 1 mole of ${HCl}_{(g)}$ is formed.

Solution

Q.
When 22.4 litres of $H_{2 (g)}$ is mixed with 11.2 litres of ${Cl}_{2 (g)}$ , each at STP, the moles of ${HCl}_{(g)}$ formed is equal to [2014]

1 1 mol of ${HCl}_{(g)}$

2 2 mol of ${HCl}_{(g)}$

3 0.5 mol of ${HCl}_{(g)}$

4 1.5 mol of ${HCl}_{(g)}$

Ans.
(1)

1 mole $\equiv$ 22.4 litres at STP.

$n_{H_{2}} = \frac{22.4}{22.4} = 1 mol; n_{{Cl}_{2}} = \frac{11.2}{22.4} = 0.5 mol$

Reaction is as,

Here, $C l_{2}$ is the limiting reagent. So, 1 mole of ${HCl}_{(g)}$ is formed.

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Solution

Q. When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g), each at STP, the moles of HCl(g) formed is equal to [2014]

1 1 mol of HCl(g)

2 2 mol of HCl(g)

3 0.5 mol of HCl(g)

4 1.5 mol of HCl(g)

Ans. (1) 1 mole ≡ 22.4 litres at STP. nH2=22.422.4=1 mol; nCl2=11.222.4=0.5 mol Reaction is as, Here, Cl2 is the limiting reagent. So, 1 mole of HCl(g) is formed.

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Q.
When 22.4 litres of $H_{2 (g)}$ is mixed with 11.2 litres of ${Cl}_{2 (g)}$ , each at STP, the moles of ${HCl}_{(g)}$ formed is equal to [2014]

1 1 mol of ${HCl}_{(g)}$

2 2 mol of ${HCl}_{(g)}$

3 0.5 mol of ${HCl}_{(g)}$

4 1.5 mol of ${HCl}_{(g)}$

Ans.
(1)

1 mole $\equiv$ 22.4 litres at STP.

$n_{H_{2}} = \frac{22.4}{22.4} = 1 mol; n_{{Cl}_{2}} = \frac{11.2}{22.4} = 0.5 mol$

Reaction is as,

Here, $C l_{2}$ is the limiting reagent. So, 1 mole of ${HCl}_{(g)}$ is formed.