How much hydrogen will be released when an excess of metallic sodium is applied to 15 g of propanol.

| August 9, 2021 | education

Let’s execute the solution:
1. Let’s write the equation:
2С3Н7ОН + 2Na = 2C3H7ONa + H2 – substitution reaction, hydrogen and sodium propionate are released;
2. We calculate the molar masses of substances:
M (C3H7OH) = 60 g / mol;
M (H2) = 2 g / mol.
3. Determine the amount of moles of propanol by the formula:
Y (C3H7OH) = m / M = 15/60 = 0.25 mol;
According to the condition of the problem, sodium is taken in excess, which means that the calculation is made by the number of moles of alcohol С3Н7ОН.
4. We make up the proportion:
0.25 mol (C3H7OH) – X mol (H2);
– 2 mol – 1 mol from here, X mol (H2) = 0.25 * 1/2 = 0.125 mol.
5. Find the volume of hydrogen:
V (H2) = 0.125 * 22.4 = 2.8 liters.
Answer: during the reaction, 2.8 liters of hydrogen was released.



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