Question

ammonia (nh₃) gas and oxygen (o₂) gas react to form nitrogen (n₂) gas and water (h₂o) vapor. suppose you have 7.0 mol of nh₃ and 11.0 mol of o₂ in a reactor.

| could half the nh₃ react? | ∘ yes
∘ no |

if you answered yes, calculate how many moles of n₂ would be produced after half the nh₃ was used up. round your answer to the nearest 0.1 mol.

mol

Explanation:

Step1: Balance the chemical equation

The reaction between \( \text{NH}_3 \) and \( \text{O}_2 \) is: \( 4\text{NH}_3 + 3\text{O}_2
ightarrow 2\text{N}_2 + 6\text{H}_2\text{O} \).

Step2: Determine moles of \( \text{NH}_3 \) reacting

Half of \( 7.0 \, \text{mol} \) of \( \text{NH}_3 \) is \( \frac{7.0}{2}= 3.5 \, \text{mol} \) of \( \text{NH}_3 \) reacting.

Step3: Find moles of \( \text{O}_2 \) required

From the balanced equation, the mole ratio of \( \text{NH}_3 \) to \( \text{O}_2 \) is \( 4:3 \). Let \( x \) be moles of \( \text{O}_2 \) required for \( 3.5 \, \text{mol} \) of \( \text{NH}_3 \).
\( \frac{4}{3}=\frac{3.5}{x} \), solving for \( x \): \( x = \frac{3.5\times3}{4}= 2.625 \, \text{mol} \) of \( \text{O}_2 \). We have \( 11.0 \, \text{mol} \) of \( \text{O}_2 \), which is more than \( 2.625 \, \text{mol} \), so half the \( \text{NH}_3 \) can react (answer is yes).

Step4: Calculate moles of \( \text{N}_2 \) produced

Mole ratio of \( \text{NH}_3 \) to \( \text{N}_2 \) is \( 4:2 = 2:1 \). For \( 3.5 \, \text{mol} \) of \( \text{NH}_3 \), moles of \( \text{N}_2 \) produced: \( \frac{2}{4}\times3.5 = 1.75 \approx 1.8 \, \text{mol} \) (rounded to nearest 0.1).

Answer:

yes, 1.8