When an excess of metallic sodium acted on a mixture containing 6.2 g of ethylene glycol and an unknown

When an excess of metallic sodium acted on a mixture containing 6.2 g of ethylene glycol and an unknown mass of glycerin, 5.6 liters of hydrogen were released. Calculate the percentage of the mixture.

m is the mass of the substance
v – amount of substance
Mr – molecular weight
V is the volume of the substance
22.4 – molar volume of gases

1. Make up the equation of reactions and place the coefficients.
2. Calculate the amount of hydrogen released in two reactions. To do this, you need to use the formula v = V / 22.4.
v (H2 total) = 5.6 / 22.4 = 0.25 mol.
3. Calculate the amount of substance for ethylene glycol according to the formula v = m / Mr.
Mr (ethylene glycol) = 62 g / mol.
v (ethylene glycol) = 6.2 / 62 = 0.1 mol
4. Looking at the reaction equation, it can be seen that the amount of ethylene glycol is equal to the amount of the released hydrogen, since the coefficients in front of these substances are equal to one.
This means that during the reaction of sodium with ethylene glycol, 0.1 mol of hydrogen was released.
And the remaining 0.15 mol is in the reaction with glycerol.
v (H2 in the reaction with glycerol) = 0.25 – 0.1 = 0.15 mol.
5. To calculate the amount of a substance of glycerin, it is necessary to make a proportion of the coefficients in front of the substances (hydrogen and glycerin) and the amount of substances in moles.
2 = 3 (row of coefficients)
x = 0.15 (string of moles)
x = 0.15 * 2/3 = 0.1 mol
v (glycerol) = 0.1 mol.
6. Find the mass of glycerin by the formula m = Mr * v.
Mr (glycerin) = 92 g / mol
m (glycerin) = 92 * 0.1 = 9.2 g
7. The total mass of the mixture of glycerin and ethylene glycol:
m (mixture) = 9.2 + 6.2 = 15.4 g
8. Mass fraction of glycerin:
w (glycerin) = 9.2 / 15.4 * 100 = 60%
w (ethylene glycol) = 100 – 60 = 40%