Answer to Question #111584 in Chemistry for andrea boni

Question

CH​ (g) + 2O​ (g) → CO​ (g) + 2H​ O(g), if 15.00 L of CH​ is burned with 4​ 2​ 2​ 2​ 4​
excess O​ at 23°C and 1.05 atm, what mass of H​ O is produced

Answer 1

P = 1.05 atm;

V = 15.00 L;

T = 23^oC = (23^oC + 273.15) = 296.15 K;

R = 0.08206 L atm K^-1 mol^-1.

Solution:

The balanced chemical equation is:

CH​₄(g) + 2O​₂(g) → CO​₂(g) + 2H​₂O(g)

Let's use the ideal gas equation to find the moles of CH₄.

The ideal gas equation can be expressed as: PV = nRT.

n = PV / RT

Then,

n(CH₄) = (1.05 atm × 15.00 L) / (0.08206 L atm K^-1 mol^-1 × 296.15 K) = 0.6481 mol

n(CH₄) = 0.6481 mol.

According to the chemical equation: n(CH₄) = n(H₂O)/2

n(H₂O) = 2 × n(CH₄) = 2 × 0.6481 mol = 1.2962 mol.

Moles of H₂O = Mass of H₂O / Molar mass of H₂O

Molar mass of H₂O = M(H₂O) = 18.01528 g/mol.

Finally, we can find the number of grams of H₂O:

n(H₂O) = m(H₂O) / M(H₂O);

m(H₂O) = n(H₂O) × M(H₂O);

m(H₂O) = (1.2962 mol × 18.01528 g/mol) = 23.3514 g = 23.35 g

Mass of H₂O = m(H₂O) = 23.35 g.

Answer: 23.35 grams of H₂O is produced.