Question

Please answer this question.. With explanation...

What is the volume of N2O(g) formed at STP when 2 moles of each KOH and NO react as per following reaction?

2KOH + 4NO -------> 2KNO2 + N2O + H2O

(1) 5.6
(2) 22.4L
(3) 44.8 L
(4) 11.2 L
​

Answer 1

2mol KOH produce N2O=1mol

so 1mol KOH produce N2O=1/2mol

no. of mole of N2O=mass of N2O/22.4

1/2=mass of N2O/22.4

mass of N2O=1/2×22.4

=11.2 L

HOPE IT WILL HELP YOU!!!!!!!!!!!!!!

please mark me as brainliest

Answer 2

Answer: (4) 11.2 L

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

$2KOH+4NO\rightarrow KNO_2+N_2O+H_2O$

According to the stochiometry of the equation:

4 moles of $NO$ react with 2 moles of $KOH$

Thus 2 moles of $NO$ will react with =$\frac{ 2}{4}\times 2=1$ mole of $KOH$

Thus $NO$ is a limiting reagent as it limits the formation of product and $KOH$ is the excess reagent.

4 moles of $NO$ give 1 mole of $N_2O$ gas.

Thus 2 moles of $NO$ will give =$\frac{1 }{4}\times 2=\frac{1}{2}$ moles of $N_2O$ gas.

1 mole of a gas at STP occupy = 22.4 L of $N_2O$ gas

$\frac{1}{2}$ mole of gas will occupy=$\frac{22.4}{1}\times\frac{1}{2}=11.2L$ of $N_2O$  gas.