Question

In the reaction of 100. g of AgNO3 with 100. g of K2CrO4 how much Ag2CrO4 will be produced, and which is the excess reagent?

• 100. g of Ag2CrO4, AgNO3

• 28.5 g of Ag2CrO4, K2CrO4

• 124 g of Ag2CrO4, AgNO3

• 97.7 g of Ag2CrO4, AgNO3

• 171 g of Ag2CrO4, K2CrO4

Answer 1

Consider the balanced equation :
2 AgNO3 + K2CrO4 ---> Ag2CrO4(s) + 2 KNO3(Aq)

The net ionic equation :

2 Ag+ (aq) + CrO42- (aq)   ------------> Ag2CrO4 (s)

Here, 2 mole of AgNO3 reacts with 1 mole of K2CrO4 to produce 1 mole of Ag2CrO4.

Given, mass of AgNO3= 100 g.

Mass of K2CrO4= 100 g.

Molar mass of AgNO3= 170 g/ml

Molar mass of K2CrO4= 194 g/ml.

Calculating no. Of moles:

No. Of moles of AgNO3 = 170/100= 1.7 mole.

No. Of moles of K2CrO4= 194/100= 1.94 mole.

From the equation, we know that 1 mole of K2CrO4 reacts with 2 mole of AgNO3.

Therefore, 1.94 g of K2CrO4 reacts with 3.88 mole of AgNO3, but only 1.74 mole of AgNO3 is present.

Therefore, AgNO3 is the limiting reagent and K2CrO4 is the excess reagent.

Now, 2 mole AgNO3. --------->. 1 mole of K2CrO4.

1.70 mole AgNO3 . ----------> 1.70×(1/2) = 0.85 mole of K2CrO4.

Yeild of Ag2CrO4= 0.85× 331.73. (Mole × molar mass).

= 281.97 g.

Hope that it helps.