Question

In: Chemistry

29.7 mL of 4.6 M Sodium Phosphate is mixed with 72.0 mL of 4.0 M Calcium...

29.7 mL of 4.6 M Sodium Phosphate is mixed with 72.0 mL of 4.0 M Calcium Bromide to form a precipitate. What precipitate is formed in this reaction? Calculate the theoretical mass (in grams) of the precipitate using only the volume and molarity of the Sodium Phosphate.Calculate the theoretical mass (in grams) of the precipitate using only the volume and molarity of the Calcium Bromide.What is the theoretical yield (in grams) of the precipitate?What is the concentration (in M) of the reactant in excess?

Expert Solution

A precipitate of calcium phosphate is obtained. The balanced chemical equation is shown below.

29.7 mL of 4.6 M Sodium Phosphate corresponds to

0.137 moles of sodium phosphate will give calcium phosphate.

The molar mass of calcium phosphate is 310.2 g/mol

Theoretical mass (in grams) of the precipitate using only the volume and molarity of the Sodium Phosphate is

72.0 mL of 4.0 M Calcium Bromide corresponds to

0.288 moles of Calcium Bromide will give calcium phosphate.

The molar mass of calcium phosphate is 310.2 g/mol

Theoretical mass (in grams) of the precipitate using only the volume and molarity of the Calcium Bromide is

Calcium Bromide is excess reagent and Sodium Phosphate is limiting reagent.

The theoretical yield (in grams) of the precipitate is 21.2 g.

0.137 moles of sodium phosphate will react with of calcium bromide.

of calcium bromide will remain.

Total volume

The concentration (in M) of the reactant in excess

queen_honey_blossom answered 2 years ago

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