What volume of hydrogen will be released if 1.15 ml of absolute alcohol (p = 0.8 g / ml) reacted with 2 mol of sodium?

Given:
V solution (C2H5OH) = 1.15 ml
ρ solution (С2Н5ОН) = 0.8 g / ml
n (Na) = 2 mol
To find:
V (H2) -?
Decision:
1) Write the reaction equation:
2C2H5OH + 2Na => 2C2H5OHa + H2 ↑;
2) In the problem, we need the molar mass of С2Н5ОН, since Na is already given in the form of an amount of substance, and Н2 is a gas (in the formula for finding the volume of gas, the molar mass is not used). Using the periodic system of D.I.Mendeleev, make a calculation that will be numerically equal to the molecular weight:
Mr (C2H5OH) = Ar (C) * 2 + Ar (H) * 6 + Ar (O) = 12 * 2 + 1 * 6 + 16 = 46 g / mol;
3) Calculate the mass of the C2H5OH solution:
m solution (С2Н5ОН) = ρ solution (С2Н5ОН) * V solution (С2Н5ОН) = 0.8 * 1.15 = 0.92 g;
4) Since according to the condition absolute alcohol is given (practically free of water), the water content in it can be neglected. So we can find the mass of С2Н5ОН:
m (C2H5OH) = m solution (C2H5OH) = 0.92 g;
5) Determine the amount of substance С2Н5ОН:
n (C2H5OH) = m (C2H5OH) / Mr (C2H5OH) = 0.92 / 46 = 0.02 mol;
6) When comparing the amount of C2H5OH and Na, it turns out that the amount of C2H5OH is in short supply, so the calculations will be carried out for it;
7) Given the reaction equation, the amount of substance H2 should be calculated:
n (H2) = n (C2H5OH) / 2 = 0.02 / 2 = 0.01 mol;
8) Find the volume of the released H2 gas:
V (H2) = n (H2) * Vm = 0.01 * 22.4 = 0.224 l.
Answer: The volume of released H2 is 0.224 liters.