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alexandr402 [8]
4 years ago
14

Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi

ng to the following equation:
3H2(g) + N2(g) → 2NH3(g)

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
L

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
L
Chemistry
2 answers:
Westkost [7]4 years ago
6 0
1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
 

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2

Second, convert the number of moles to liters of gas at STP:

3.75 mol * 22.4 L/mol =  84 liters of H2

Answer: 84 liters

 </span>



vaieri [72.5K]4 years ago
4 0

the answer is

125 & 84

the second part is

0.38

0.38

27

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How many dm3 of hydrogen are released when 3 g of potassium is reacted with hydroiodic acid?
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Answer:

V = 0.0859dm³

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<em>Moles H₂:</em>

0.0767 moles of K * (1mol H₂ / 2mol K) = 0.03836 moles H₂

Using: PV = nRT; V = nRT / P

<em>Where V is volume in dm³,</em>

<em>n are moles of gas: 0.03836 moles,</em>

<em>R is gas constant = 0.082atm*dm³/molK</em>

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<em>and P is pressure = 1atm</em>

The volume of the gas is:

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<h3>V = 0.0859dm³</h3>

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