# To 50 g of a 6% aqueous solution of potassium stearate was added 20 g of a 1% sulfuric acid solution

To 50 g of a 6% aqueous solution of potassium stearate was added 20 g of a 1% sulfuric acid solution. Will gas evolve when sodium carbonate is added to the separated aqueous solution?

Let’s write down the reaction equations:
2С17H35COOK + H2SO4 = 2С17H35COOH + K2So4
So that when adding sodium carbonate, there must be an acid in the remaining solution, if the acid all reacts, then when adding sodium carbonate there will be nothing.
Find the mass of pure potassium stearate:
w (in-va) = m (in-va) * 100% / m (solution)
m (in-va) = w (in-va) * m (solution) / 100%
m (С17H35COOK) = w (С17H35COOK) * m (solution) / 100% = 6 * 50/100 = 3 g.
Find the mass of pure acid:
m (H2SO4) = w (H2SO4) * m (solution) / 100% = 1 * 20/100 = 0.2 g.
It can be seen from the reaction equation that:
ν (С17H35COOK) / 2 = ν (H2SO4)
m (С17H35COOK) / 2 * M (С17H35COOK) = m (H2SO4) / M (H2SO4)
Let’s define the molar masses:
M (С17H35COOK) = 12 * 18 + 1 * 35 + 16 * 2 + 39 = 322 g / mol
M (H2SO4) = 1 * 2 + 32 + 16 * 4 = 98 g / mol
Determine the mass of acid required for the potassium stearate to react:
m (H2SO4) = m (С17H35COOK) * M (H2SO4) / 2 * M (С17H35COOK)
Substitute the numerical values:
m (H2SO4) = m (С17H35COOK) * M (H2SO4) / 2 * M (С17H35COOK) = 3 * 98/2 * 322 = 0.45 g.
For the reaction, 0.45 g of acid is needed, we have less of it, which means that it will all react and there is no acid for sodium carbonate.
Answer: gas will not be released, the acid has completely reacted with potassium stearate.

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