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how many grams of lead (ii) iodide will be precipitated by adding sufficient Pb(NO3)2 to trace with 630mL of 0.268M Ki solution?
2KI(aq)+Pb(NO3)2(aq) → PbI2+2KNO3(aq)​

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The mass of PbI2 produced from a trace amount of 630mL of 0.268M KI solution is 39.2 g.

The equation of the reaction is; 2KI(aq)+Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq)​

From the question, number of moles of KI = 630/1000 L × 0.268M = 0.169 moles

Now;

2 moles of KI is required to precipitate 1 mole of PbI2

0.169 moles of KI is required to precipitate 0.169 moles × 1 mole/2 moles

= 0.085 moles

Mass of PbI2 = Number of moles of PbI2 × Molar mass

Mass of PbI2 = 0.085 moles × 461.01 g/mol

Mass of PbI2 = 39.2 g

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