Determine the mass of water in the neutralization reaction between the sulfuric acid solution

| September 3, 2021 | education

Determine the mass of water in the neutralization reaction between the sulfuric acid solution and 250 grams of sodium hydroxide solution.

To solve the problem, we write down the given: m (NaOH) = 250 g. The sodium hydroxide solution reacted with sulfuric acid.
Find the mass of water that was formed as a result of the reaction.
Solution:
Let’s write down the reaction equation.
NaOH + H2SO4 = Na2SO4 + H2O
Let’s arrange the coefficients and get:
2NaOH + H2SO4 = Na2SO4 + 2H2O
Let’s calculate the molar masses of sodium hydroxide and water.
M (NaOH) = 23 + 16 + 1 = 40 g / mol
M (H2O) = 1 * 2 + 16 = 18 g / mol
Since sodium hydroxide and water entered 2 mol, it is necessary to multiply the molar masses by 2. we get:
M (NaOH) = 80 g
M (H2O) = 36 g
We write 250 g over sodium hydroxide, and under it we write 80 g. Over water we write x g, and under it – 36 g.
Let’s compose and solve the proportion.
x = 250 * 36/80 = 112.5 g
Answer: m (H2O) = 112.5 g



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