When 10 g of sodium hydroxide interacts with sulfuric acid, salt and water are formed. Find the mass of these substances

| September 21, 2021 | education

Let’s execute the solution:
1. In accordance with the condition of the problem, we compose the equation:
2NaOH + H2SO4 = Na2SO4 + 2H2O – ion exchange, obtained sodium sulfate and water;
2. Calculations:
M (NaOH) = 39.9 g / mol.
M (Na2SO4) = 141.8 g / mol.
M (H2O) = 18 g / mol.
3. Determine Y (mol) of the starting material:
Y (NaOH) = m / M = 10 / 39.9 = 0.25 mol.
Y (H2O) = 0.25 mol since the amount of substances according to the equation is equal to 2 mol.
4. Proportion:
0.25 mol (NaOH) – X mol (Na2SO4);
-2 mol -1 mol
Hence, X mol (Na2SO4) = 0.25 * 1/2 = 0.125 mol.
5. Find the mass of products:
m (H2O) = Y * M = 0.25 * 18 = 4.5 g.
m (Na2SO4) = Y * M = 0.125 * 141.8 = 17.73 g.
Answer: the mass of sodium sulfate is 17.73 g, and water is 4.5 g.



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