Question

How many grams of Hydronium chromate are produced when 43.4 g of Tin (IV) chromate combines with35.2 g of Hydronium hydrogen phosphate? Use the following balanced equation:2 (H3O)2HPOA + 1 Sn(CrO4)2 ---> 2 (H30)2CrOA + 1 Sm(HPOA)2

Answer 1

Mass of (H30)2CrO = 38 g

Step-by-step explanation

Given:

Mass of Sn(CrO4)2 = 43.4 g

Mass of (H3O)2HPO4 = 35.2 g

Required: The mass of (H30)2CrO4 that will be produced

Solution:

Calculate the possible mass that could be produced by each reactant, so as to determine the limiting reagent. Use stoichiometry.

For Sn(CrO4)2:

`\begin{gathered} 43.4\text{ g Sn\lparen CrO}_4\text{\rparen}_2\text{ x }\frac{1\text{ mole Sn\lparen CrO}_4)_2}{350.70\text{ g Sn\lparen CrO}_4)_2}\text{ x }\frac{2\text{ mole \lparen H}_3\text{0\rparen}_2\text{CrO}_4}{1\text{ mole Sn\lparen CrO}_4)_2}\text{ x }\frac{153.9\text{ g \lparen H}_3O)_2CrO_4}{1\text{ mol \lparen H}_3O)_2CrO_4} \\ \\ =\text{ 38 g \lparen H}_3\text{O\rparen}_2\text{CrO}_4 \end{gathered}`

For (H3O)2HPO4

`\begin{gathered} 35.2\text{ g \lparen H}_3\text{O\rparen}_2\text{HPO}_4\text{ x }\frac{1\text{ mole \lparen H}_3\text{O\rparen}_2\text{HPO}_4\text{ }}{133.97\text{ g }(H_3O)_2HPO_4}\text{ x }\frac{2\text{ mole}}{2\text{ mole}}\text{ x }\frac{153.9\text{ g \lparen H}_3\text{O\rparen}_2\text{CrO}_4}{1\text{ mole \lparen H}_3\text{O\rparen}_2\text{CrO}_4} \\ \\ =\text{ 40.43 g \lparen H}_3\text{O\rparen}_2\text{CrO}_4 \end{gathered}`

Sn(CrO4)2 will produce less (H30)2CrO4 therefore, Sn(CrO4)2 is the limiting reagent.