Calculate the mass of sodium acetate CH3COONa and sodium hydroxide that will be required to obtain methane

Calculate the mass of sodium acetate CH3COONa and sodium hydroxide that will be required to obtain methane with a volume of 56 liters under normal conditions.

Let’s implement the solution:
1. We write down the equation according to the problem statement:
X g -? X g -? V = 56 liters.
CH3COONa + NaOH = CH4 + Na2CO3 – exchange occurs when heated, methane is released;
2. Calculations:
M (CH3COONa) = 81.9 g / mol;
M (NaOH) = 39.9 g / mol;
M (CH4) = 16 g / mol.
3. Proportion:
1 mol of gas at normal level – 22.4 liters;
X mol (CH4) – 56 liters. hence, X mol (CH4) = 1 * 56 / 22.4 = 2.5 mol;
Y (CH3COONa) = 2.5 mol;
Y (NaOH) = 2.5 mol.
4. Find the masses of the starting materials:
m (CH3COONa) = Y * M = 2.5 * 81.9 = 204.75 g;
m (NaOH) = Y * M = 2.5 * 39.9 = 99.75 g.
Answer: to carry out the process, sodium acetate weighing 204.75 g and soda lime – 99.75 g are required.