Calculate the masses of ethyl alcohol and metallic sodium required to obtain 1.12 liters of hydrogen.

1. Let’s compose the reaction equation:

2 С2Н5ОН + 2 Na = 2 С2Н5ОNa + H2

According to the equation:

C2H5OH and Na are taken in an amount of 2 mol;
H2 formed 1 mol, i.e. 22.4 l
Let’s find the mass of С2Н5ОН and Na by the formula:

m (C2H5OH) = n * M = 2 * (12 * 2 + 1 * 5 + 16 +1) = 2 * 46 = 92 g

m (Na) = n * M = 2 mol * 23 g / mol = 46 g

2. Let’s calculate the mass of С2Н5ОН, making up the proportion:

x g С2Н5ОН – 1.12 l H2

92 g С2Н5ОН – 22.4 l H2

Hence, x = 92 * 1.12 / 22.4 = 4.6 g

3. Let’s calculate the mass of Na, making up the proportion:

xg Na – 1.12 L H2

46 g Na – 22.4 L H2

Hence, x = 46 * 1.12 / 22.4 = 2.3 g.