To aluminum carbide was added 400 ml of water, while a gas with a volume of 13.44 liters was evolved.

To aluminum carbide was added 400 ml of water, while a gas with a volume of 13.44 liters was evolved. 160 g of 30% sodium hydroxide solution was added to the resulting mixture. determine the mass fractions of substances in the resulting solution.

Given:
V (H2O) = 400 ml
V (gas) = ​​13.44 l
m solution (NaOH) = 160 g
ω (NaOH) = 30%

Find:
ω (in-in) -?

1) Al4C3 + 12H2O => 4Al (OH) 3 + 3CH4 ↑;
Al (OH) 3 + NaOH => Na [Al (OH) 4];
2) m (H2O) = ρ * V = 1 * 400 = 400 g;
3) n (H2O) = m / M = 400/18 = 22.222 mol;
4) n (CH4) = V / Vm = 13.44 / 22.4 = 0.6 mol;
5) n (Al (OH) 3) = n (CH4) * 4/3 = 0.6 * 4/3 = 0.8 mol;
6) m (NaOH) = ω * m solution / 100% = 30% * 160/100% = 48 g;
7) n (NaOH) = m / M = 48/40 = 1.2 mol;
8) n (Na [Al (OH) 4]) = n (Al (OH) 3) = 0.8 mol;
9) m (Na [Al (OH) 4]) = n * M = 0.8 * 118 = 94.4 g;
10) n react. (NaOH) = n (Al (OH) 3) = 0.8 mol;
11) n rest. (NaOH) = n (NaOH) – n re. (NaOH) = 1.2 – 0.8 = 0.4 mol;
12) m rest. (NaOH) = n rest. * M = 0.4 * 40 = 16 g;
13) n (Al4C3) = n (CH4) / 3 = 0.6 / 3 = 0.2 mol;
14) m (Al4C3) = n * M = 0.2 * 144 = 28.8 g;
15) m (CH4) = n * M = 0.6 * 16 = 9.6 g;
16) m solution = m (Al4C3) + m (H2O) – m (CH4) + m solution (NaOH) = 28.8 + 400 – 9.6 + 160 = 579.2 g;
17) ω (Na [Al (OH) 4]) = m * 100% / m solution = 94.4 * 100% / 579.2 = 16.3%;
18) ω rest. (NaOH) = m rest. * 100% / m p-ra = 16 * 100% / 579.2 = 2.76%.

Answer: The mass fraction of Na [Al (OH) 4] is 16.3%; NaOH – 2.76%.