Question

ammonium phosphate $\left(\left(\mathrm{nh}_4\
ight)_3\mathrm{po}_4\
ight)$ is an important ingredient in many fertilizers. it can be made by reacting phosphoric acid $\left(\mathrm{h}_3\mathrm{po}_4\
ight)$ with ammonia $\left(\mathrm{nh}_3\
ight)$.
what mass of ammonium phosphate is produced by the reaction of 1.4 g of ammonia?
be sure your answer has the correct number of significant digits.
\boxed{} g

Explanation:

Step1: Write the balanced reaction

The reaction between phosphoric acid ($\ce{H_3PO_4}$) and ammonia ($\ce{NH_3}$) to form ammonium phosphate ($\ce{(NH_4)_3PO_4}$) is:
$$\ce{H_3PO_4 + 3NH_3 -> (NH_4)_3PO_4}$$

Step2: Calculate moles of $\ce{NH_3}$

Molar mass of $\ce{NH_3}$: $14.01 + 3(1.008) = 17.034\ \text{g/mol}$.
Moles of $\ce{NH_3} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.4\ \text{g}}{17.034\ \text{g/mol}} \approx 0.0822\ \text{mol}$.

Step3: Relate moles of $\ce{NH_3}$ to $\ce{(NH_4)_3PO_4}$

From the balanced equation, $3\ \text{mol}\ \ce{NH_3}$ produces $1\ \text{mol}\ \ce{(NH_4)_3PO_4}$.
Moles of $\ce{(NH_4)_3PO_4} = \frac{0.0822\ \text{mol}\ \ce{NH_3}}{3} \approx 0.0274\ \text{mol}$.

Step4: Calculate molar mass of $\ce{(NH_4)_3PO_4}$

Molar mass: $3(14.01 + 4(1.008)) + 30.97 + 4(16.00) = 3(18.042) + 30.97 + 64.00 = 54.126 + 30.97 + 64.00 = 149.096\ \text{g/mol}$.

Step5: Calculate mass of $\ce{(NH_4)_3PO_4}$

Mass = moles × molar mass = $0.0274\ \text{mol} × 149.096\ \text{g/mol} \approx 4.08\ \text{g}$.
Rounding to 2 significant digits (since 1.4 g has 2 sig figs), the mass is $4.1\ \text{g}$ (or more precisely, $4.0\ \text{g}$ if strict, but rechecking:

Wait, recalculating moles of $\ce{NH_3}$: $\frac{1.4}{17.034} \approx 0.08219\ \text{mol}$.
Moles of $\ce{(NH_4)_3PO_4} = 0.08219 / 3 \approx 0.027397\ \text{mol}$.
Mass: $0.027397 × 149.096 ≈ 4.08\ \text{g}$. With 2 sig figs, $4.1\ \text{g}$ (or 4.0 g if 1.4 is 2 sig figs, but 1.4 has two, so 4.1 or 4.0? Wait, let's do exact:

$1.4\ \text{g}\ \ce{NH_3} × \frac{1\ \text{mol}\ \ce{NH_3}}{17.03\ \text{g}\ \ce{NH_3}} × \frac{1\ \text{mol}\ \ce{(NH_4)_3PO_4}}{3\ \text{mol}\ \ce{NH_3}} × \frac{149.1\ \text{g}\ \ce{(NH_4)_3PO_4}}{1\ \text{mol}\ \ce{(NH_4)_3PO_4}}$

$= 1.4 × \frac{149.1}{17.03 × 3} ≈ 1.4 × \frac{149.1}{51.09} ≈ 1.4 × 2.918 ≈ 4.09\ \text{g} ≈ 4.1\ \text{g}$ (2 sig figs).

Answer:

$\boxed{4.1}$ (or 4.0, depending on rounding; but 1.4 has 2 sig figs, so 4.1 is appropriate)