QuestionMay 20, 2025

4. From the balanced chemical equation solve the following questions bellow: N_(2)+3H_(2)arrow 2NH_(3) a) Determine the number moles of ammonia (NH_(3)) produced from the complete reaction of 0.75 moles of hydrogen gas (H_(2)) [5pts] __ __ __ b) Determine the number of grams of nitrogen gas (N_(2)) required to produce 550 g of ammonia (NH_(3)) [5pt

4. From the balanced chemical equation solve the following questions bellow: N_(2)+3H_(2)arrow 2NH_(3) a) Determine the number moles of ammonia (NH_(3)) produced from the complete reaction of 0.75 moles of hydrogen gas (H_(2)) [5pts] __ __ __ b) Determine the number of grams of nitrogen gas (N_(2)) required to produce 550 g of ammonia (NH_(3)) [5pt
4. From the balanced chemical equation solve the following questions bellow: N_(2)+3H_(2)arrow 2NH_(3) a) Determine the number moles of ammonia (NH_(3)) produced from the complete reaction of 0.75 moles of hydrogen gas (H_(2)) [5pts] __ __ __ b) Determine the number of grams of nitrogen gas (N_(2)) required to produce 550 g of ammonia (NH_(3)) [5pt

Solution
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Answer

a) 0.5 moles of NH_3 ### b) 452.9 grams of N_2 Explanation 1. Calculate moles of NH_3 from H_2 Use the mole ratio from the equation: 3 moles of H_2 produce 2 moles of NH_3. Therefore, 0.75 \text{ moles of } H_2 \times \frac{2 \text{ moles of } NH_3}{3 \text{ moles of } H_2} = 0.5 \text{ moles of } NH_3. 2. Calculate moles of NH_3 from grams Molar mass of NH_3 is 14 + 3 \times 1 = 17 \text{ g/mol}. Moles of NH_3 = \frac{550 \text{ g}}{17 \text{ g/mol}} = 32.35 \text{ moles}. 3. Calculate moles of N_2 required Use the mole ratio from the equation: 1 mole of N_2 produces 2 moles of NH_3. Therefore, 32.35 \text{ moles of } NH_3 \times \frac{1 \text{ mole of } N_2}{2 \text{ moles of } NH_3} = 16.175 \text{ moles of } N_2. 4. Calculate grams of N_2 Molar mass of N_2 is 2 \times 14 = 28 \text{ g/mol}. Grams of N_2 = 16.175 \text{ moles} \times 28 \text{ g/mol} = 452.9 \text{ g}.

Explanation

1. Calculate moles of $NH_3$ from $H_2$<br /> Use the mole ratio from the equation: 3 moles of $H_2$ produce 2 moles of $NH_3$. Therefore, $0.75 \text{ moles of } H_2 \times \frac{2 \text{ moles of } NH_3}{3 \text{ moles of } H_2} = 0.5 \text{ moles of } NH_3$.<br /><br />2. Calculate moles of $NH_3$ from grams<br /> Molar mass of $NH_3$ is $14 + 3 \times 1 = 17 \text{ g/mol}$. Moles of $NH_3 = \frac{550 \text{ g}}{17 \text{ g/mol}} = 32.35 \text{ moles}$.<br /><br />3. Calculate moles of $N_2$ required<br /> Use the mole ratio from the equation: 1 mole of $N_2$ produces 2 moles of $NH_3$. Therefore, $32.35 \text{ moles of } NH_3 \times \frac{1 \text{ mole of } N_2}{2 \text{ moles of } NH_3} = 16.175 \text{ moles of } N_2$.<br /><br />4. Calculate grams of $N_2$<br /> Molar mass of $N_2$ is $2 \times 14 = 28 \text{ g/mol}$. Grams of $N_2 = 16.175 \text{ moles} \times 28 \text{ g/mol} = 452.9 \text{ g}$.
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