Question

according to the following reaction, how many moles of ammonium nitrite are necessary to form 0.697 moles water? ammonium nitrite (aq)→nitrogen (g)+water (l) moles ammonium nitrite use the references to access important values if needed for this question. 3 item attempts remaining

Explanation:

Step1: Write the balanced chemical equation

The balanced equation for the decomposition of ammonium nitrite is $NH_4NO_2(aq)
ightarrow N_2(g)+2H_2O(l)$.

Step2: Determine the mole - ratio

From the balanced equation, the mole - ratio of ammonium nitrite to water is $n_{NH_4NO_2}:n_{H_2O}=1:2$.

Step3: Calculate moles of ammonium nitrite

We know that $n_{H_2O} = 0.697$ moles. Using the mole - ratio $\frac{n_{NH_4NO_2}}{n_{H_2O}}=\frac{1}{2}$, we can solve for $n_{NH_4NO_2}$. So $n_{NH_4NO_2}=\frac{1}{2}n_{H_2O}$.
Substitute $n_{H_2O} = 0.697$ moles into the equation: $n_{NH_4NO_2}=\frac{1}{2}\times0.697$ moles.
$n_{NH_4NO_2}=0.3485$ moles.

Answer:

$0.349$ moles (rounded to three significant figures)