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2. Nitrogen gas and hydrogen gas combine to produce ammonia gas (NH3). What

volume of hydrogen gas at STP would be required to react with 10.0 g of nitrogen
gas?

User Hirse
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1 Answer

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Answer:

The balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia gas is:

N2 + 3H2 → 2NH3

From the equation, we can see that 3 moles of hydrogen gas are required to react with 1 mole of nitrogen gas to produce 2 moles of ammonia gas.

To find the volume of hydrogen gas required to react with 10.0 g of nitrogen gas at STP, we can use the following steps:

1. Convert the mass of nitrogen gas to moles:

moles of N2 = mass / molar mass

moles of N2 = 10.0 g / 28.0 g/mol

moles of N2 = 0.357 mol

2. Use the mole ratio from the balanced equation to find the moles of hydrogen gas required:

moles of H2 = 3 x moles of N2

moles of H2 = 3 x 0.357 mol

moles of H2 = 1.07 mol

3. Use the ideal gas law to find the volume of hydrogen gas at STP:

PV = nRT

where P = 1 atm, V is the volume we want to find, n = moles of gas, R = 0.0821 L·atm/mol·K (the gas constant), and T = 273 K (standard temperature at STP).

Rearranging the equation to solve for V, we get:

V = nRT / P

V = 1.07 mol x 0.0821 L·atm/mol·K x 273 K / 1 atm

V = 24.5 L

Therefore, the volume of hydrogen gas required to react with 10.0 g of nitrogen gas at STP is 24.5 L.

User Leighanne
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