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?

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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


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