Question

for the formation of 3.65g of HCL gas what volume of hydrogen gas and chlorine gas at stp are required

Answer 1

Answer: 1.12 L of hydrogen gas and  1.12 L of chlorine gas are required.

Explanation: Moles of HCl =$\frac{\text{ given mass}}{\text{ molar mass}}= \frac{3.65g}{36.5g/mole}=0.1moles$

$\frac{1}{2}H_2+\frac{1}{2}Cl_2\rightarrow HCl$

From the given balanced equation:

1 mole of HCl require =$\frac{1}{2}$ mole of Hydrogen and $\frac{1}{2}$ ole of chlorine

0.01 mole of HCl require=0.05 mole of Hydrogen and 0.05 mole of chlorine.

According to Avogadro's law, 1 mole of gas occupies=22.4 L of gas at STP.

Thus 0.05 mole of every gas occupies=$\frac{22.4}{1}\times 0.05=1.12$ Litres at STP.