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 6.4 g of ammonia? be sure your answer has the correct number of significant digits. g ×10⁰

Explanation:

Step1: Balance the chemical equation

The reaction is \( \ce{4NH_3 + 5O_2 -> 4NO + 6H_2O} \).

Step2: Calculate moles of \( \ce{NH_3} \)

Molar mass of \( \ce{NH_3} \) is \( 14.01 + 3\times1.008 = 17.034 \, \text{g/mol} \). Moles of \( \ce{NH_3} = \frac{6.4 \, \text{g}}{17.034 \, \text{g/mol}} \approx 0.3757 \, \text{mol} \).

Step3: Relate moles of \( \ce{NH_3} \) to \( \ce{NO} \)

From balanced equation, \( 4 \, \text{mol} \, \ce{NH_3} \) produces \( 4 \, \text{mol} \, \ce{NO} \), so moles of \( \ce{NO} = 0.3757 \, \text{mol} \).

Step4: Calculate mass of \( \ce{NO} \)

Molar mass of \( \ce{NO} \) is \( 14.01 + 16.00 = 30.01 \, \text{g/mol} \). Mass of \( \ce{NO} = 0.3757 \, \text{mol} \times 30.01 \, \text{g/mol} \approx 11.27 \, \text{g} \), rounded to two significant digits (since 6.4 has two) is \( 11 \, \text{g} \) (or more accurately, checking significant digits: 6.4 has two, so result should have two. Wait, 6.4 is two sig figs. Let's recalculate moles: \( 6.4 / 17.034 \approx 0.3757 \), moles of NO same, mass: \( 0.3757 \times 30.01 \approx 11.27 \), which rounds to 11 g? Wait, no, 6.4 is two sig figs, so 11 g (two sig figs) or maybe 11. g? Wait, let's do exact calculation:

\( \frac{6.4 \, \text{g} \, \ce{NH_3}}{17.03 \, \text{g/mol}} \times \frac{4 \, \text{mol} \, \ce{NO}}{4 \, \text{mol} \, \ce{NH_3}} \times 30.01 \, \text{g/mol} \, \ce{NO} = \frac{6.4 \times 30.01}{17.03} \approx \frac{192.064}{17.03} \approx 11.28 \, \text{g} \), which with two sig figs is \( 11 \, \text{g} \) (or 1.1×10¹ g). Wait, maybe I made a mistake in sig figs: 6.4 is two, so the answer should have two. So approximately 11 g. But let's check again. Wait, 6.4 is two significant digits. So the calculation:

Moles of NH3: 6.4 / 17.0 = 0.376 (approx, 17.0 is three sig figs, but 6.4 is two, so moles is 0.38? Wait, no, when dividing, the number of sig figs is determined by the least, so 6.4 (two) / 17.034 (four) = two sig figs: 0.38 mol? Wait, no, 6.4 / 17.034 = 0.3757, which is ~0.38 when rounded to two sig figs? Wait, no, 0.3757 is 0.38 (two sig figs). Then moles of NO is 0.38 mol, mass: 0.38 × 30.01 = 11.4038, which rounds to 11 g (two sig figs). So the mass is approximately 11 g.

Answer:

\boxed{11}