Question

ammonium phosphate \\( ( ( nh _ { 4 } ) _ { 3 } po _ { 4 } ) \\) is an important ingredient in many fertilizers. it can be made by reacting phosphoric acid \\( ( h _ { 3 } po _ { 4 } ) \\) with ammonia \\( ( nh _ { 3 } ) \\).
what mass of ammonium phosphate is produced by the reaction of 9.1 g of phosphoric acid?
be sure your answer has the correct number of significant digits.
\\( \square g \\)

Explanation:

Step1: Write the balanced chemical equation

$$\ce{H3PO4 + 3NH3=(NH4)3PO4}$$

Step2: Calculate the molar mass of \(H_3PO_4\)

The molar mass of \(H = 1\space g/mol\), \(P=31\space g/mol\), \(O = 16\space g/mol\).
\(M_{H_3PO_4}=3\times1 + 31+4\times16=98\space g/mol\)

Step3: Calculate the moles of \(H_3PO_4\)

\(n_{H_3PO_4}=\frac{m}{M}=\frac{9.1\space g}{98\space g/mol}\approx0.0929\space mol\)

Step4: Determine the mole ratio

From the balanced equation, the mole ratio of \(H_3PO_4\) to \((NH_4)_3PO_4\) is \(1:1\). So \(n_{(NH_4)_3PO_4}=n_{H_3PO_4} = 0.0929\space mol\)

Step5: Calculate the molar mass of \((NH_4)_3PO_4\)

The molar mass of \(N = 14\space g/mol\), \(H = 1\space g/mol\), \(P=31\space g/mol\), \(O = 16\space g/mol\).
\(M_{(NH_4)_3PO_4}=3\times(14 + 4\times1)+31+4\times16=149\space g/mol\)

Step6: Calculate the mass of \((NH_4)_3PO_4\)

\(m_{(NH_4)_3PO_4}=n\times M=0.0929\space mol\times149\space g/mol\approx13.8\space g\)

Answer:

\(14\space g\)