Question

2nh3(g) → n2(g) + 3h2(g) the reaction for the decomposition of ammonia (nh3) can be written as shown. if a student starts with 21.7 g of nh3, how many grams of hydrogen (h2) gas will be produced by the reaction? 1.28 g 32.5 g 2.55 g 3.85 g

Answer 1

2NH3 --> N2 + 3H2
17*2g. 28g. 3*2g

Thus 34g of NH3 gives 6g of H2

Then 21.7g of NH3 will give 21.7*6/34 = 3.81g

Hence option D should be correct

Answer 2

Answer : 3.828 grams of hydrogen gas will be produced by the reaction.

Solution : Given,

Mass of $NH_3$ = 21.7 g

Molar mass of $NH_3$ = 17 g/mole

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the moles of $NH_3$ gas.

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=\frac{21.7g}{17g/mole}=1.276moles$

• The given balanced chemical reaction is,

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

From the reaction, we conclude that

2 moles of $NH_3$ gas react to gives 3 moles of $H_2$ gas

1.276 moles of $NH_3$ gas react to gives $\frac{3}{2}\times 1.276=1.914$ moles of $H_2$ gas

Now we have to calculate the mass of $H_2$ gas.

Mass of $H_2$ gas = Moles of $H_2$ × Molar mass of $H_2$

Mass of $H_2$ gas = 1.914 g × 2 g/mole = 3.828 g

Therefore, 3.828 grams of hydrogen gas will be produced by the reaction.