Question

CH4 (g) + H2O(g) → CO(g) + 3H2 (g) What is the volume of steam, H2O(g), needed to completely react with 53.50L of methane gas at a constant pressure and temperature? What is the volume of carbon monoxide, CO, produced? What is the volume of Hydrogen gas, H2, produced? What is the total volume of gas produced?

Answer 1

Answer: 185.5 L.

Explanation: To solve this question, we need to use the stoichiometry of the chemical reaction to determine the amounts of each gas produced. The balanced chemical equation is:

CH4 (g) + H2O(g) → CO(g) + 3H2(g)

According to the stoichiometry of the reaction, 1 mole of methane reacts with 1 mole of water to produce 1 mole of carbon monoxide and 3 moles of hydrogen. Therefore, we can use the ideal gas law to calculate the volumes of each gas produced.

Given that the initial volume of methane gas is 53.50 L, we can first calculate the number of moles of methane present using the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the pressure and temperature are constant, we can write:

n = PV/RT

where R = 0.08206 L atm/K mol is the gas constant.

n(CH4) = (1 atm)(53.50 L)/(0.08206 L atm/K mol)(298 K) = 2.23 mol

This means that 2.23 moles of methane react with 2.23 moles of water to produce 2.23 moles of carbon monoxide and 6.69 moles of hydrogen.

To determine the volume of water needed to react with all the methane, we can use the stoichiometry of the reaction again:

1 mol CH4 reacts with 1 mol H2O

Therefore, the number of moles of water required is also 2.23 mol. We can calculate the volume of water using the ideal gas law:

n(H2O) = PV/RT

V(H2O) = n(H2O)RT/P = (2.23 mol)(0.08206 L atm/K mol)(298 K)/(1 atm) = 46.4 L

Therefore, the volume of steam required to react with all the methane is 46.4 L.

Next, we can use the stoichiometry of the reaction to calculate the volumes of carbon monoxide and hydrogen produced:

1 mol CH4 produces 1 mol CO

1 mol CH4 produces 3 mol H2

Therefore, the number of moles of carbon monoxide and hydrogen produced are also 2.23 mol and 6.69 mol, respectively. We can calculate their volumes using the ideal gas law:

V(CO) = n(CO)RT/P = (2.23 mol)(0.08206 L atm/K mol)(298 K)/(1 atm) = 46.4 L

V(H2) = n(H2)RT/P = (6.69 mol)(0.08206 L atm/K mol)(298 K)/(1 atm) = 139.1 L

Therefore, the volume of carbon monoxide produced is 46.4 L and the volume of hydrogen produced is 139.1 L.

The total volume of gas produced is the sum of the volumes of carbon monoxide and hydrogen:

V(total) = V(CO) + V(H2) = 46.4 L + 139.1 L = 185.5 L

Therefore, the total volume of gas produced is 185.5 L.

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Answer 2

To solve this problem, we need to use stoichiometry to relate the volume of methane to the volumes of steam, carbon monoxide, and hydrogen produced.

The balanced chemical equation is:

CH4 (g) + H2O(g) → CO(g) + 3H2 (g)

The stoichiometric ratio of steam to methane is 1:1, so the volume of steam needed is also 53.50 L.

To determine the volumes of carbon monoxide and hydrogen gas produced, we need to use the stoichiometric coefficients in the balanced equation. For every 1 mole of methane consumed, 1 mole of steam is consumed, 1 mole of carbon monoxide is produced, and 3 moles of hydrogen gas are produced.

First, we need to convert the volume of methane gas to moles using the ideal gas law:

PV = nRT

n = PV/RT

n = (1 atm) x (53.50 L) / [(0.08206 L·atm/mol·K) x (298 K)]

n = 2.189 mol

Therefore, 2.189 moles of methane react with 2.189 moles of steam to produce 2.189 moles of carbon monoxide and 6.567 moles of hydrogen gas.

To convert the moles of each gas to volume, we use the ideal gas law again:

V = nRT/P

For carbon monoxide:

n = 2.189 mol

R = 0.08206 L·atm/mol·K

T = 298 K

P = 1 atm

V = (2.189 mol) x (0.08206 L·atm/mol·K) x (298 K) / (1 atm)

V = 53.68 L

For hydrogen gas:

n = 6.567 mol

R = 0.08206 L·atm/mol·K

T = 298 K

P = 1 atm

V = (6.567 mol) x (0.08206 L·atm/mol·K) x (298 K) / (1 atm)

V = 160.76 L

The total volume of gas produced is the sum of the volumes of carbon monoxide and hydrogen gas:

53.68 L + 160.76 L = 214.44 L

Therefore, the volume of steam needed is 53.50 L, the volume of carbon monoxide produced is 53.68 L, the volume of hydrogen gas produced is 160.76 L, and the total volume of gas produced is 214.44 L.