Question 2 (1 point) A student needs to produce 45.18 grams of nitrogen gas (N_(2)) for a lab experiment. How many grams of ammonia (NH_(3)) will be required to produce this amount of nitrogen? Round your answer to the nearest 0.01 and remember to include units and substance! NH_(3)... N_(2)+H_(2) Your Answer: square square I Answer units Question 3(1 point)

## Step1: Write the balanced chemical equation<br />### The given reaction is unbalanced. To balance it, we write:<br />\[<br />2NH_{3} \rightarrow N_{2} + 3H_{2}<br />\]<br />This shows that 2 moles of ammonia ($NH_{3}$) produce 1 mole of nitrogen gas ($N_{2}$).<br /><br />## Step2: Calculate the molar masses of $NH_{3}$ and $N_{2}$<br />### Using atomic masses from the periodic table:<br />- Molar mass of $NH_{3}$: $14.01 + (3 \times 1.008) = 17.034 \, \text{g/mol}$<br />- Molar mass of $N_{2}$: $2 \times 14.01 = 28.02 \, \text{g/mol}$<br /><br />## Step3: Determine the moles of $N_{2}$ required<br />### The mass of $N_{2}$ needed is $45.18 \, \text{g}$. Using the formula $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$:<br />\[<br />\text{Moles of } N_{2} = \frac{45.18}{28.02} \approx 1.612 \, \text{mol}<br />\]<br /><br />## Step4: Relate moles of $NH_{3}$ to moles of $N_{2}$<br />### From the balanced equation, 2 moles of $NH_{3}$ produce 1 mole of $N_{2}$. Therefore:<br />\[<br />\text{Moles of } NH_{3} = 2 \times \text{Moles of } N_{2} = 2 \times 1.612 = 3.224 \, \text{mol}<br />\]<br /><br />## Step5: Calculate the mass of $NH_{3}$ required<br />### Using the formula $\text{mass} = \text{moles} \times \text{molar mass}$:<br />\[<br />\text{Mass of } NH_{3} = 3.224 \times 17.034 \approx 54.89 \, \text{g}<br />\]