Question

ammonia reacts with oxygen gas to form nitrogen monoxide and water. if 5.59 moles ammonia react, how many moles of oxygen are needed to react in stoichiometric proportions? (hint: you need to write a balanced equation for the reaction first.) mol o₂ what mass of water will be produced? g h₂o

Explanation:

Step1: Write the balanced chemical equation

The reaction is \(4NH_{3}+5O_{2}
ightarrow4NO + 6H_{2}O\)

Step2: Calculate moles of \(O_{2}\)

From the balanced equation, the mole ratio of \(NH_{3}\) to \(O_{2}\) is \(4:5\).
If \(n(NH_{3}) = 5.59\space mol\), then \(n(O_{2})=\frac{5}{4}\times n(NH_{3})\)
\(n(O_{2})=\frac{5}{4}\times5.59 = 6.99\space mol\)

Step3: Calculate moles of \(H_{2}O\)

From the balanced equation, the mole ratio of \(NH_{3}\) to \(H_{2}O\) is \(4:6 = 2:3\)
\(n(H_{2}O)=\frac{3}{2}\times n(NH_{3})\)
\(n(H_{2}O)=\frac{3}{2}\times5.59 = 8.385\space mol\)

Step4: Calculate mass of \(H_{2}O\)

The molar mass of \(H_{2}O\) is \(M = 18.015\space g/mol\)
\(m(H_{2}O)=n(H_{2}O)\times M\)
\(m(H_{2}O)=8.385\times18.015\approx151.1\space g\)

Answer:

\(6.99\space mol\space O_{2}\)
\(151.1\space g\space H_{2}O\)