How many grams of potassium iodide will produce 500 grams of silver iodide, when there is an excess of silver nitrate? Why?

A. 235

B. 707g

C. 166g

D. 301g

None of the options are correct. The answer to the question is 353.2g

Explanation:

The equation for the reaction between is given below:

KI + AgNO3 - > KNO3 + AgI

Molar Mass of KI = 39 + 127 = 166g/mol

Molar Mass of AgI = 108 + 127 = 235g/mol

From the equation,

166g of KI produced 235g of AgI.

Therefore, Xg of KI will produce 500g of AgI i. e

Xg of KI = (166 x 500) / 235 = 353.2g