Question

chemistry experts solve this question but reply fast and I will mark it as brainliest .. For the formation of 3.65g of HCL gas , what volume of hydrogen gas and chlorine gas are required at NTP condition .

Answer 1

Hope u understand this.

Answer 2

Answer : The volume of hydrogen gas and chlorine gas required at NTP condition are, 1.12 L and 1.12 L respectively.

Solution : Given,

Mass of HCl = 3.65 g

Molar mass of HCl = 36.5 g/mole

First we have to calculate the moles of HCl gas.

$\text{Moles of HCl}=\frac{\text{Given mass of HCl}}{\text{Molar mass of HCl}}=\frac{3.65g}{36.5g/mole}=0.1moles$

Now we have to calculate the moles of hydrogen gas and chlorine gas.

The balanced chemical reaction will be,

$H_2(g)+CL_2(g)\rightarrow 2HCl(g)$

From the balanced reaction, we conclude that

2 moles of HCl gas formed from 1 mole of hydrogen gas

0.1 moles of HCl gas formed from $\frac{0.1}{2}=0.05$ moles of hydrogen gas

and,

2 moles of HCl gas formed from 1 mole of chlorine gas

0.1 moles of HCl gas formed from $\frac{0.1}{2}=0.05$ moles of chlorine gas

Now we have to calculate the volume hydrogen gas and chlorine gas are required at NTP condition.

At NTP, 1 mole contains 22.4 L volume of gas

As, 1 mole of contains 22.4 L volume of hydrogen gas

So, 0.05 mole contains $22.4\times 0.05=0.12L$ volume of hydrogen gas.

and,

As, 1 mole of contains 22.4 L volume of chlorine gas

So, 0.05 mole contains $22.4\times 0.05=0.12L$ volume of chlorine gas.

Therefore, the volume of hydrogen gas and chlorine gas required at NTP condition are, 1.12 L and 1.12 L respectively.