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 Mio?

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 mio?

Penjelasan:

mmol La³⁺ = 50 × 0.25 = 12.5

mmol IO₃⁻ = 75 × 0.302 = 22.65

find limiting reactant:

$\frac{mmol \: {La}^{3 + } {mula}^{2} }{coef \: {La}^{3 + } } = \frac{12.5}{1} = 12.5$

$\frac{mmol \: {IO₃}^{ - } {mula}^{2} }{coef \: {IO₃}^{ - } } = \frac{22.65}{3} = 7.55$

7.55 < 12.5 so IO₃⁻ = limiting reactant (reactant that is totally consumed when the chemical reaction is completed.)

chemical reaction is in the pic:

mass of La(IO₃)₃ = mmol × Mr = 7.55 × 663.6 = 5010.18 mg

= 5.01018 gr

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 mio?