50.0 mL of a solution that is 0.0800 M in $Ba X^{2 +}$ and 0.0600 M in $Sr X^{2 +}$ is treated with excess $(NH X_{4}) X_{2} CO X_{3}$ in the presence of $NH X_{4} Cl$ and $NH X_{4} OH$ . Assuming complete precipitation of both ions as carbonates, find the total mass of precipitate in grams. Atomic masses: Ba 137, Sr 88, C 12, O 16. Give the answer to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 1.23 g

Moles of

Ba X^{2 +}

0.0500 \times 0.0800 = 4.00 \times 1 0^{- 3}

and of

Sr X^{2 +}

3.00 \times 1 0^{- 3}

. Molar masses are

BaCO X_{3}

= 197 and

SrCO X_{3}

= 148 g/mol. Mass =

4.00 \times 1 0^{- 3} \times 197 + 3.00 \times 1 0^{- 3} \times 148 = 1.232

g, i.e. 1.23 g.