Aluminum carbide weighing 2.16 g was treated with excess water, and the resulting gas was burned

Aluminum carbide weighing 2.16 g was treated with excess water, and the resulting gas was burned in excess of oxygen. The combustion products were completely absorbed in a solution of 10.66 ml with a mass fraction of NaOH 20% (p = 1.22 g / ml). Determine the mass fractions of substances in the solution.

Let us write down all the equations of the course of the reactions.
1.Al4C3 + 12H2O = 4Al (OH) 3 + 3CH4.
The resulting gas is methane.
2.CH4 + 2O2 = CO2 + 2H2O.
Combustion products are water and carbon dioxide.
3. NaOH + Н2О = NaOH • Н2О.
NaOH + CO2 = NaHCO3.
Formation of hydrate and bicarbonate.

Find the number of moles of aluminum carbide:
n (Al4C3) = m / M = 2.16 / 143.96 = 0.015 mol.
According to the reaction equation, from 1 mol of Al4C3, 3 mol of CH4 is formed, which means:
n (CH4) = 3 • n (Al4C3) = 3 • 0.015 = 0.045 mol.
n (CO2) = n (CH4) = 0.045 mol.
N (H2O) = 2 * n (CH4) = 2 * 0.045 = 0.09 mol.
N (NaHCO3) = n (CO2) = 0.045 mol.
N (NaOH • H2O) = n (H2O) = 0.09 mol.

Knowing the number of moles, we find the mass of the substance:
M1 (NaHCO3) = n • M = 0.046 • 84 = 3.78 g.
M2 (NaOH • H2O) = 0.09 • 58 = 5.22 g.
M (NaOH) = V • p = 10.66 • 1.22 = 13.0 g.
M3 (NaOH) = m • W / 100% = 13 • 0.2 = 2.6 g.
M (solution) = m1 + m2 + m3 = 3.78 + 5.22 + 2.6 = 11.6 g.

The mass fractions of substances in the solution will accordingly be equal:
W (NaHCO3) = m1 / m (solution) • 100% = 3.78 / 11.6 • 100% = 32.59%.
W (NaHCO3) = m2 / m (solution) • 100% = 5.22 / 11.6 • 100% = 45%.

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