Question

the volume of ammonia obtained by the combination of 10mL of N2 and 30mL H2 is...
Explained it.
A) 20mL B) 40mL C) 30mL D) 10mL

Answer 1

here is the answer
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Answer 2

Answer: 20 ml

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given volume}}{\text{Molar volume}}$  

For $N_2$

Given volume= 10 ml

Putting values in above equation, we get:

$\text{Moles of}N_2 =\frac{10}{22.4}=0.00045mol$

For $H_2$

Given volume= 30 ml

Putting values in above equation, we get:

$\text{Moles of}H_2 =\frac{30}{22400}=0.00135mol$

According to avogadro's law, 1 mole of every substance occupies 22400 ml and contains avogadro's number $6.023\times 10^{23}$ of particles.

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

1 mole of $N_2$ reacts with 3 moles of $H_2$

Thus 0.00045 moles of $N_2$ reacts with =$\frac{3}{1}\times 0.00045=0.00135$ moles of $H_2$

Thus the reactants combine completely and both act as limiting reagents.

1 mole of $N_2$ gives = 2 moles of $NH_3$

Thus 0.00045 moles of $N_2$ give =$\frac{2}{1}\times 0.00045=0.0009$ moles of $NH_3$

Volume of $NH_3=moles\times {\text {molar volume}}=0.0009mol\times 22400ml=20ml$

Thus the volume of ammonia obtained by the combination of 10mL of $N_2$ and 30mL $H_2$ is 20 ml.