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Part B How many grams of NH3 can be produced from 3.10 mol of N2 and excess H2? Express your answer numerically in grams. Answer: To determine the grams of NH3 produced, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced equation is: N2 + 3H2 -> 2NH3 From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the molar ratio between N2 and NH3 is 1:2. Given that we have 3.10 mol of N2, we can use the molar ratio to calculate the moles of NH3 produced: 3.10 mol N2 * (2 mol NH3 / 1 mol N2) = 6.20 mol NH3 To convert the moles of NH3 to grams, we need to use the molar mass of NH3, which is 17.03 g/mol. Therefore, the grams of NH3 produced is: 6.20 mol NH3 * 17.03 g/mol = 105.646 g NH3 Therefore, 105.646 grams of NH3 can be produced from 3.10 mol of N2 and excess H2. Part C How many grams of H2 are needed to produce 14.69 g of NH3? Express your answer numerically. Answer: To determine the grams of H2 needed, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced equation is: N2 + 3H2 -> 2NH3 From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the molar ratio between H2 and NH3 is 3:2. Given that we have 14.69 g of NH3, we can use the molar mass of NH3 to calculate the moles of NH3: 14.69 g NH3 * (1 mol NH3 / 17.03 g NH3) = 0.863 mol NH3 To convert the moles of NH3 to moles of H2, we can use the molar ratio between H2 and NH3: 0.863 mol NH3 * (3 mol H2 / 2 mol NH3) = 1.2955 mol H2 To convert the moles of H2 to grams, we need to use the molar mass of H2, which is 2.02 g/mol. Therefore, the grams of H2 needed is: 1.2955 mol H2 * 2.02 g/mol = 2.615 g H2 Therefore, 2.615 grams of H2 are needed to produce 14.69 g of NH3.

          Part B
How many grams of NH3 can be produced from 3.10 mol of N2 and excess H2? Express your answer numerically in grams. 
Answer:
To determine the grams of NH3 produced, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced equation is:

N2 + 3H2 -> 2NH3

From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the molar ratio between N2 and NH3 is 1:2.

Given that we have 3.10 mol of N2, we can use the molar ratio to calculate the moles of NH3 produced:

3.10 mol N2 * (2 mol NH3 / 1 mol N2) = 6.20 mol NH3

To convert the moles of NH3 to grams, we need to use the molar mass of NH3, which is 17.03 g/mol. Therefore, the grams of NH3 produced is:

6.20 mol NH3 * 17.03 g/mol = 105.646 g NH3

Therefore, 105.646 grams of NH3 can be produced from 3.10 mol of N2 and excess H2.

Part C
How many grams of H2 are needed to produce 14.69 g of NH3? Express your answer numerically. 
Answer:
To determine the grams of H2 needed, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced equation is:

N2 + 3H2 -> 2NH3

From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the molar ratio between H2 and NH3 is 3:2.

Given that we have 14.69 g of NH3, we can use the molar mass of NH3 to calculate the moles of NH3:

14.69 g NH3 * (1 mol NH3 / 17.03 g NH3) = 0.863 mol NH3

To convert the moles of NH3 to moles of H2, we can use the molar ratio between H2 and NH3:

0.863 mol NH3 * (3 mol H2 / 2 mol NH3) = 1.2955 mol H2

To convert the moles of H2 to grams, we need to use the molar mass of H2, which is 2.02 g/mol. Therefore, the grams of H2 needed is:

1.2955 mol H2 * 2.02 g/mol = 2.615 g H2

Therefore, 2.615 grams of H2 are needed to produce 14.69 g of NH3.

Added by Daniel B.

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