Determine the amount of sodium chloride substance, the volume of carbon monoxide (4) and the mass of water

| May 13, 2021 | education

Determine the amount of sodium chloride substance, the volume of carbon monoxide (4) and the mass of water, which are formed by the interaction of 79.5 g of sodium carbonate with an excess of hydrochloric acid.

1. Let’s calculate the amount of sodium carbonate substance according to the formula n = m: M, where M is the molar mass (we find it according to the periodic table, adding the masses of atoms in the formula).

2. We make the equation of the reaction between sodium carbonate and hydrochloric acid.

3. Determine the quantitative ratio of substances according to the equation.

a) When 1 mol of sodium carbonate interacts with 2 mol of hydrochloric acid, 2 mol of sodium chloride NaCl are formed, that is, the substances are in a ratio of 1: 2. Then the amount of NaCl will be 2 times more than that of sodium carbonate.

n (NaCl) = 2 x 0.75 mol = 1.5 mol.

b) Sodium carbonate and water are in a 1: 1 ratio. Their amounts of matter will be the same. n (water) = 0.75 mol. Knowing n (water), you can find the mass by the formula:

m = nM, m = 0.75 x 18 g / mol = 13.5 g.

c) Sodium carbonate and carbon dioxide react 1: 1.

n (gas) = ​​0.75 mol.

Let’s calculate the volume using the formula:

V = Vn n, where Vn is the molar volume equal to 22.4 l / mol, and n is the amount of substance.

V = 22.4 L / mol x 0.75 mol = 16.8 L.

Answer: V (gas) = ​​16.8 l; n (NaCl) = 1.5 mol; m (water) = 13.5 g.



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