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Standard XII

Chemistry

Ideal Gas Equation

10 moles of a...

Question

$10$ moles of a mixture of hydrogen and oxygen gases at a pressure of $1$ atm at a constant volume and temperatures, react to form $3.6 g$ of liquid water. What will be the pressure of the resulting mixture?

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Solution

$2 H_{2} + O_{2} ⟶ 2 H_{2} O$

Initially Xmol (10-X)mol 0

After reaction ( X-a) (10-X- $\frac{a}{2}$ ) a

$a = \frac{3.6}{18} = 0.2$

$∴$ resulting mole of gases in the mixture:

$X - 0.2 + 10 - X - \frac{0.2}{2}$

$= 10 - 0.3 = 9.7$

$∴$ from relation,

$\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}$

$\frac{1}{10} = \frac{P_{2}}{9.7}$

$P_{2} = 0.97 a t m .$

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10 moles of a mixture of hydrogen and oxygen gases at a pressure of 1 atm at a constant volume and temperatures, react to form 3.6g of liquid water. What will be the pressure of the resulting mixture?

2H2+O2⟶2H2OInitially Xmol (10-X)mol 0After reaction ( X-a) (10-X-a2) aa=3.618=0.2∴ resulting mole of gases in the mixture:X−0.2+10−X−0.22=10−0.3=9.7∴ from relation,P1n1=P2n2 110=P29.7P2=0.97atm.

$10$ moles of a mixture of hydrogen and oxygen gases at a pressure of $1$ atm at a constant volume and temperatures, react to form $3.6 g$ of liquid water. What will be the pressure of the resulting mixture?