Question

How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? __ Ba(NO3)2 + __ Na2SO4 __ BaSO4 + __ NaNO3

Answer 1

Answer:

2.91 grams of slid barium sulfate will be formed.

Explanation:

Moles of barium nitrate:

$0.50 M=\frac{Moles}{0.025 L}$

Moles of sulfuric acid = 0.0125 moles

$Ba(NO_3)_2 +Na_2SO_4\rightarrow BaSO_4 +2NaNO_3$

According to reaction , 1 mol of barium nitrate gives 1 mol of barium sulfate

Then 0.0125 moles of barium nitrate gives:

$0.0125\times 1=0.0125 mol$ of sodium

Mass of barium sulfate =$0.0125 mol\times 233.38 g/mol=2.91 g$

2.91 grams of slid barium sulfate will be formed.