N2(g) + 3H2(g) ==> 2NH3(g)
A. If 16.99 g of hydrogen is reacted at STP, volume of ammonia is produced?
B. If 35.82 L of nitrogen is reacted at STP, what mass of hydrogen must also be reacted?
C. At STP, what volume of hydrogen is needed to produce 70.9 L of ammonia;
D. What volume of nitrogen is needed to produce 0.947 mol of ammonia?
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To solve these questions, we need to use the balanced chemical equation and the concept of stoichiometry. Let's solve each question step by step:
A. If 16.99 g of hydrogen is reacted at STP, the first thing we need to do is convert the given mass of hydrogen to moles. The molar mass of hydrogen (H2) is approximately 2 g/mol. So, we divide the given mass by the molar mass to obtain the number of moles:
16.99 g H2 / 2 g/mol = 8.495 mol H2
According to the balanced equation, 3 moles of hydrogen react with 2 moles of ammonia. So, we can set up a ratio to find the number of moles of ammonia produced:
3 mol H2 : 2 mol NH3 = 8.495 mol H2 : x mol NH3
Now, we can solve for x by cross-multiplication:
3x = (2/3) * 8.495
x = (2/3) * 8.495 / 3
x ≈ 4.497 mol NH3
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. Therefore, 4.497 moles of ammonia will occupy:
4.497 mol NH3 * 22.4 L/mol = 100.647 L
So, approximately 100.647 L of ammonia is produced.
B. If 35.82 L of nitrogen is reacted at STP, we need to find the mass of hydrogen needed. Firstly, let's convert the given volume of nitrogen to moles. At STP, 1 mole of gas occupies 22.4 L. So, the number of moles of nitrogen is:
35.82 L N2 / 22.4 L/mol = 1.599 mol N2
From the balanced equation, we know that it takes 3 moles of hydrogen to react with 1 mole of nitrogen. Therefore, we can set up the following ratio:
3 mol H2 : 1 mol N2 = x mol H2 : 1.599 mol N2
Solving for x, we get:
x = 3 * 1.599 / 1
x ≈ 4.797 mol H2
Now, let's convert the moles of hydrogen to grams using the molar mass of hydrogen (approximately 2 g/mol):
4.797 mol H2 * 2 g/mol = 9.594 g H2
Therefore, approximately 9.594 grams of hydrogen must also be reacted.
C. At STP, what volume of hydrogen is needed to produce 70.9 L of ammonia? Similar to the previous questions, we'll start by finding the number of moles of ammonia produced. The given volume of ammonia can be converted to moles using the molar volume at STP:
70.9 L NH3 / 22.4 L/mol = 3.166 mol NH3
Using the stoichiometry of the reaction, 3 moles of hydrogen react with 2 moles of ammonia. Setting up the ratio:
3 mol H2 : 2 mol NH3 = x mol H2 : 3.166 mol NH3
Solving for x, we get:
x = (3 * 3.166) / 2
x ≈ 4.749 mol H2
Now, we can convert the moles of hydrogen to volume at STP:
4.749 mol H2 * 22.4 L/mol = 106.377 L
Therefore, approximately 106.377 L of hydrogen is needed.
D. To find the volume of nitrogen needed to produce 0.947 mol of ammonia, we'll use a similar approach. Firstly, convert the given moles of ammonia to moles of nitrogen using the stoichiometry:
0.947 mol NH3 * (1 mol N2 / 2 mol NH3) = 0.4735 mol N2
Now, we can convert the moles of nitrogen to volume at STP using the molar volume:
0.4735 mol N2 * 22.4 L/mol = 10.59 L
Therefore, approximately 10.59 L of nitrogen is needed to produce 0.947 mol of ammonia.