Question

Given the following chemical equation, determine how many grams of N2 are produced by 8.49 g of H2O2 and 5.72 g of N2H4.

2H2O2(l) + N2H4(l) ---> 4H20(g) + N2(g)

Answer 1

Answer on Question#40485-Chemistry-Other

Question

Given the following chemical equation: $2\mathrm{H}_2\mathrm{O}_2(\mathrm{l}) + \mathrm{N}_2\mathrm{H}_4(\mathrm{l}) \rightarrow 4\mathrm{H}_2\mathrm{O}(\mathrm{g}) + \mathrm{N}_2(\mathrm{g})$

Determine how many grams of $\mathsf{N}_2$ are produced by $8.49\mathrm{g}$ of $\mathsf{H}_2\mathsf{O}_2$ and $5.72\mathrm{g}$ of $\mathsf{N}_2\mathsf{H}_4$.

Solution

$\mathrm{M(H_2O)} = 18\mathrm{g / mol},\mathrm{M(N_2H_4)} = 32\mathrm{g / mol},\mathrm{M(N_2)} = 28\mathrm{g / mol}.$

Number of moles of the reactants:

$$\begin{array}{l} n \left(H _ {2} O\right) = m \left(H _ {2} O\right) / M \left(H _ {2} O\right) = 8.49 / 18 = 0.47 \mathrm{mol} \\ n \left(N _ {2} H _ {4}\right) = m \left(N _ {2} H _ {4}\right) / M \left(N _ {2} H _ {4}\right) = 5.72 / 32 = 0.18 \mathrm{mol} \\ \end{array}$$

The actual molar ratio of the reactants:

$$n \left(\mathrm {H} _ {2} \mathrm {O}\right) / n \left(\mathrm {N} _ {2} \mathrm {H} _ {4}\right) = 0.47 / 0.18 = 2.61 / 1$$

As is clear from the chemical equation the theoretical molar ratio $n(H_2O) / n(N_2H_4) = 2 / 1$.

So, water is taken in excess and some of it remains unreacted. That is why the mass of $\mathsf{N}_2$ produced must be calculated based on the amount of $\mathsf{N}_2\mathsf{H}_4$ not $\mathsf{H}_2\mathsf{O}$.

As is clear from the chemical equation the molar ratio $n(N_2) / n(N_2H_4) = 1 / 1$, i.e. $n(N_2) = 0.18$ mol.

Mass of $\mathsf{N}_2$ produced: $m(\mathsf{N}_2) = n(\mathsf{N}_2)\cdot \mathsf{M}(\mathsf{N}_2) = 0.18\cdot 28 = 5.04\mathrm{g}$

Answer: 5.04 g