Question

What mass of solid La(IO3)3 (663.6 g/mol) is formed when 50.0 mL of 0.250 M La3+ are mixed with 75.0 mL of 0.302 M IO3-?

Answer 1

Explanation:

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Answer 2

Answer: 5.02 g of La(IO3)3

Explanation:

La3+ + IO3-  --->  La(IO3)3

La3+= 0.250 mol/L * $\frac{1 L}{1000mL}$ * 50.0 ml = 0.0125 mol La3+

IO3- = 0.302 mol/L * $\frac{1 L}{1000mL}$ * 75 .0 ml = 0.02265 mol IO3-

0.02265 mol IO3- * $\frac{1 mol La(IO3)3 }{3 mol IO3-}$ * $\frac{663.6 g La(IO3)3 }{1 mol La(IO3)3}$ = 5.02 g of La(IO3)3