Question

ammonium phosphate (nh₄)₃po₄ is an important ingredient in many fertilizers. it can be made by reacting phosphoric acid (h₃po₄) with ammonia (nh₃). what mass of ammonium phosphate is produced by the reaction of 6.9 g of ammonia? round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

$$\mathrm{H_3PO_4 + 3NH_3=(NH_4)_3PO_4}$$

Step2: Calculate the molar mass of ammonia ($\mathrm{NH_3}$) and ammonium phosphate ($\mathrm{(NH_4)_3PO_4}$)

The molar mass of $\mathrm{NH_3}$ is $M_{NH_3}=14 + 3\times1= 17\ g/mol$. The molar mass of $\mathrm{(NH_4)_3PO_4}$ is $M_{(NH_4)_3PO_4}=3\times(14 + 4\times1)+31+4\times16=149\ g/mol$.

Step3: Calculate the number of moles of ammonia

The number of moles of $\mathrm{NH_3}$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{6.9\ g}{17\ g/mol}\approx0.4059\ mol$.

Step4: Determine the mole - ratio and calculate the moles of ammonium phosphate

From the balanced equation, the mole - ratio of $\mathrm{NH_3}$ to $\mathrm{(NH_4)_3PO_4}$ is $3:1$. So the number of moles of $\mathrm{(NH_4)_3PO_4}$, $n_{(NH_4)_3PO_4}=\frac{1}{3}n_{NH_3}=\frac{1}{3}\times0.4059\ mol\approx0.1353\ mol$.

Step5: Calculate the mass of ammonium phosphate

The mass of $\mathrm{(NH_4)_3PO_4}$, $m_{(NH_4)_3PO_4}=n_{(NH_4)_3PO_4}\times M_{(NH_4)_3PO_4}=0.1353\ mol\times149\ g/mol\approx20\ g$.

Answer:

$20\ g$