How many grams of precipitate is formed if 250 g of sodium hydroxide solution with a mass fraction of sodium

| August 4, 2021 | education

How many grams of precipitate is formed if 250 g of sodium hydroxide solution with a mass fraction of sodium hydroxide of 16% are added to 48 g of magnesium sulfate?

Let’s write the reaction equation:

MgSO4 + 2NaOH → Na2SO4 + Mg (OH) 2 (sodium sulfate, magnesium hydroxide as a precipitate).

Let’s calculate the amount of magnesium sulfate substance:

n (MgSO4) = m (MgSO4) / M (MgSO4) = 48 g / 120 g / mol = 0.4 mol.

Find the mass of the solute in the sodium hydroxide solution:

mr.w. (NaOH) = (mr-pa (NaOH) * wr.h.) / 100% = (250 g * 16%) / 100% = 40 g.

Let’s calculate the amount of sodium hydroxide substance:

n (NaOH) = m (NaOH) / M (NaOH) = 40 g / 40 g / mol = 1 mol.

For 0.4 mol of magnesium sulfate, 0.8 mol of sodium hydroxide is required. Therefore, NaOH is in excess. We are calculating using MgSO4.

Let’s calculate the mass of magnesium hydroxide:

n (Mg (OH) 2) = n (MgSO4) = 0.4 mol.

m (Mg (OH) 2) = n (Mg (OH) 2) * M (Mg (OH) 2) = 0.4 mol * 58 g / mol = 23.2 g.



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