The action of 1 g of sodium per 1.1 g of a mixture of methanol and ethanol released 336 ml of hydrogen

The action of 1 g of sodium per 1.1 g of a mixture of methanol and ethanol released 336 ml of hydrogen. Calculate the mass fraction of alcohols in the mixture.

Let’s find the amount of substance H2.

n = V: Vn.

n = 0.336 L: 22.4 L / mol = 0.015 mol.

Let’s find the amount of the substance Na.

M (Na) = 23 g / mol.

n = m: M.

n (Na) = 1 g: 23 g / mol = 0.04 mol (excess).

Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

2 CH3OH + 2Na = 2 CH3ONa + H2.

2 С2H5OH + 2Na = 2 С2H5ONa + H2.

According to the reaction equation, there are 2 mol of alcohols per 1 mol of H2. The substances are in quantitative ratios of 1: 2.

n (alcohol) = 2 n (H2) = 0.015 × 2 = 0.03 mol.

Let the amount of CH3OH substance be x g, and the mass of С2H5OH y grams, then we get the expression: x + y = 0.03.

m = n × M.

M (CH3OH) = 32g / mol.

M (C2H5OH) = 46 g / mol.

The mass of CH3OH will be 32 x, and the mass of C2H5OH is 46y.

We get the expression 32 x + 46 y = 1.1.

Let’s compose a system of equations.

{x + y = 0.03,

{32 x + 46y = 1.1.

X = 0.03 – y,

32 (0.03 – y) + 46y = 1.1,

0.96 – 32y + 46y = 1.1.

0.96 – 1.1 = 32y – 46 y.

-0.14 = – 14y,

y = -0.14: 14,

y = 0.01,

x = 0.03 – 0.01,

x = 0.02.

Let us find the mass of CH3OH.

m = 0.02 × 32 = 0.64 g.

Let’s find the mass of С2H5OH.

m = 0.01 × 46 = 0.46 or (1.1 – 0.64 = 0.46).

Let’s find the mass fraction of alcohols in the mixture.

w (CH3OH) = (0.64: ​​1.1) × 100% = 58.18%.

w (C2H5OH) = (0.46: 1.1) × 100% = 41.82%.

Answer: 58.18%; 41.82%.