Calculate the mass of Na that can completely react with 16 g of methanol. Calculate the volume of H2

Calculate the mass of Na that can completely react with 16 g of methanol. Calculate the volume of H2 that will be released.

To solve it, you need to make an equation:
2Na + 2CH3OH = 2CH3ONa + H2 – substitution reaction, hydrogen gas is released;
Let’s calculate the molar masses of substances:
M (Na) = 22.9 g / mol;
M (CH3OH) = 32 g / mol;
M (H2) = 2 g / mol.
Determine the amount of moles of alcohol methanol:
Y (CH3OH) = m / M = 16/32 = 0.5 mol;
According to the reaction equation, the amount of moles of methanol and sodium are equal to 2 mol, which means that Y (Na) = 0.5 mol.
Let’s calculate the mass of sodium:
m (Na) = Y * M = 0.5 * 22.9 = 11.45 g (the substance is in short supply), taking this value into account, we make calculations.
Let’s make the proportion:
0.5 mol (Na) – X mol (H2);
-2 mol 1 mol from here, X mol (H2) = 0.5 * 1/2 = 0.25 mol.
Find the volume of hydrogen:
V (H2) = 0.25 * 22.4 = 11.2 liters.
Answer: the mass of sodium is 11.45 g, the volume of hydrogen is 11.2 liters.