Question

according to the following reaction, how many grams of ammonium nitrite are needed to form 26.0 grams of water? ammonium nitrite (aq) → nitrogen (g) + water (l) mass = grams ammonium nitrite. use the references to access important values if needed for this question. 9 more group attempts remaining. submit answer retry entire group

Explanation:

Step1: Write the balanced chemical equation

$NH_4NO_2(aq)
ightarrow N_2(g)+2H_2O(l)$

Step2: Calculate the molar mass of water and ammonium nitrite

The molar mass of $H_2O$ is $M_{H_2O}=(2\times1.01 + 16.00)\text{ g/mol}=18.02\text{ g/mol}$. The molar mass of $NH_4NO_2$ is $M_{NH_4NO_2}=(14.01+4\times1.01 + 14.01+2\times16.00)\text{ g/mol}=64.06\text{ g/mol}$

Step3: Determine the moles of water

The moles of water, $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{26.0\text{ g}}{18.02\text{ g/mol}}\approx1.443\text{ mol}$

Step4: Use the mole - ratio from the balanced equation

From the balanced equation, the mole - ratio of $NH_4NO_2$ to $H_2O$ is $1:2$. So the moles of $NH_4NO_2$, $n_{NH_4NO_2}=\frac{1}{2}n_{H_2O}=\frac{1}{2}\times1.443\text{ mol}=0.7215\text{ mol}$

Step5: Calculate the mass of ammonium nitrite

The mass of ammonium nitrite, $m_{NH_4NO_2}=n_{NH_4NO_2}\times M_{NH_4NO_2}=0.7215\text{ mol}\times64.06\text{ g/mol}\approx46.2\text{ g}$

Answer:

46.2 g