Answer to Question #175406 in General Chemistry for Anna

Question

determine the percent yield for the reaction between 6.92g of K and 4.28g of 02 if 7.36 of K20 is produced?

Answer 1

4K + O₂ = 2K₂O

Atomic mass of K = 39.0983 U

Molar mass of O₂ = 32 g/mol

Molar mass of K₂O = 94.2 g/mol

From the balanced equation it is observed that,

4 moles K react with 1 mol of O₂

ie, 4*39.0983 g of K react with 32 g of O₂

Thus, 6.92 g of K react with {(32*6.92)/(4*39.0983)} ie, 1.41 g of O₂ .

Thus, here the limiting reagent is K and the excess reagent is O₂ .

From the balanced equation it is also observed that,

From 4 moles of K, 2 moles of K₂O is produced.

ie, from (4 * 39.0983) g of K, (2 * 94.2) g of K₂O is produced.

Thus from 6.92 g of K, {(6.92*2*94.2)/(4*39.0983)} ie, 8.34 g of K₂O is produced.

ie, theoretical yield = 8.34 g

Actual yield = 7.36 g

% of yield =(actual yield/theoretical yield) *100%

= (7.36/8.34) *100%

= 88.2%