Question

ammonia (nh₃) chemically reacts with oxygen gas (o₂) to produce nitric oxide (no) and water (h₂o). what mass of nitric oxide is produced by the reaction of 8.7 g of ammonia? be sure your answer has the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

$$4NH_3 + 5O_2=4NO + 6H_2O$$

Step2: Calculate the molar mass of \(NH_3\)

The molar mass of \(N = 14.01\space g/mol\) and \(H=1.008\space g/mol\). So, \(M_{NH_3}=14.01+(3\times1.008)= 17.034\space g/mol\)

Step3: Find the number of moles of \(NH_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 8.7\space g\) and \(M = 17.034\space g/mol\). So, \(n_{NH_3}=\frac{8.7\space g}{17.034\space g/mol}\approx0.5107\space mol\)

Step4: Use mole - ratio from the balanced equation

From \(4NH_3 + 5O_2 = 4NO+6H_2O\), the mole ratio \(n_{NH_3}:n_{NO}=4:4 = 1:1\). So, \(n_{NO}=n_{NH_3}\approx0.5107\space mol\)

Step5: Calculate the molar mass of \(NO\)

The molar mass of \(N = 14.01\space g/mol\) and \(O = 16.00\space g/mol\). So, \(M_{NO}=14.01 + 16.00=30.01\space g/mol\)

Step6: Calculate the mass of \(NO\)

Using the formula \(m=n\times M\), where \(n = 0.5107\space mol\) and \(M = 30.01\space g/mol\). So, \(m_{NO}=0.5107\space mol\times30.01\space g/mol\approx15.33\space g\)

Answer:

\(15\space g\)