QuestionMay 28, 2025

Mg_(3)N_(2)(s)+6H_(2)O(l)- gt 2NH_(3)(aq)+3Mg(OH)_(2)(s) if 77.2grams of water are mixed with excess magnesium nitride, how many grams of NH_(3) are produced? square

Mg_(3)N_(2)(s)+6H_(2)O(l)- gt 2NH_(3)(aq)+3Mg(OH)_(2)(s) if 77.2grams of water are mixed with excess magnesium nitride, how many grams of NH_(3) are produced? square
Mg_(3)N_(2)(s)+6H_(2)O(l)- gt 2NH_(3)(aq)+3Mg(OH)_(2)(s) if 77.2grams of water are mixed with excess magnesium nitride, how many grams of NH_(3) are produced? square

Solution
4.5(220 votes)

Answer

24.37 grams of NH_3 are produced. Explanation 1. Calculate moles of H_2O Molar mass of H_2O is 18.02 g/mol. Moles of H_2O = \frac{77.2 \text{ g}}{18.02 \text{ g/mol}} = 4.28 \text{ mol}. 2. Determine moles of NH_3 produced From the balanced equation, 6 moles of H_2O produce 2 moles of NH_3. Thus, 4.28 \text{ mol } H_2O \times \frac{2 \text{ mol } NH_3}{6 \text{ mol } H_2O} = 1.43 \text{ mol } NH_3. 3. Calculate grams of NH_3 Molar mass of NH_3 is 17.03 g/mol. Grams of NH_3 = 1.43 \text{ mol} \times 17.03 \text{ g/mol} = 24.37 \text{ g}.

Explanation

1. Calculate moles of $H_2O$<br /> Molar mass of $H_2O$ is 18.02 g/mol. Moles of $H_2O = \frac{77.2 \text{ g}}{18.02 \text{ g/mol}} = 4.28 \text{ mol}$.<br /><br />2. Determine moles of $NH_3$ produced<br /> From the balanced equation, 6 moles of $H_2O$ produce 2 moles of $NH_3$. Thus, $4.28 \text{ mol } H_2O \times \frac{2 \text{ mol } NH_3}{6 \text{ mol } H_2O} = 1.43 \text{ mol } NH_3$.<br /><br />3. Calculate grams of $NH_3$<br /> Molar mass of $NH_3$ is 17.03 g/mol. Grams of $NH_3 = 1.43 \text{ mol} \times 17.03 \text{ g/mol} = 24.37 \text{ g}$.
Click to rate:

Similar Questions