Question

1Given the reaction how moles will be praduced many from 3.10 mol A,ass uming excess B? 2A 3B 4Y +52 =moles of Z Mole 2 For the reac tion 2 AgNO3 Na2CrOy2 Na N03 How many grams of Silver chromare, Ag 2Cr Oy are Produced from A) 29.3 g of silver nitrate, AgNO3 mass of Sjlver chromate: 9 B)

Answer 1

Ans 1.

2A + 3B(excess) →   4Y + 5Z

In the given reaction B is excess , so A is limiting reagent and it decide the mole of Z

The given reaction is practically going

2 mole of A producing mole of Z is = 5

therefore 1 mole of A producing mole of Z is = 5/2

Hence 3.10 mole of A producing mole of Z is = 5*3.10/2 = 7.75 mole

Thus 3.10 A give  7.75 mole of Z

Ans 2.

2AgNO₃ + Na₂CrO₄   → 2NaNO₃ + Ag₂CrO₄

From above reaction clear that 2 mole of AgNO₃ is give 1 mole of Ag₂CrO₄

Or 2* 169.87 gmol^-1 *mole of AgNO₃ is give 1*331.73 gmol^-1 *mole of Ag₂CrO₄

                   Note : I multiply with molar mass of compound

Or

339.74 g of AgNO₃ is give  Ag₂CrO₄ amount is = 331.73g

1 g of AgNO₃ is give  Ag₂CrO₄ amount is = 331.73g/339.74

29.3 g of AgNO₃ is give  Ag₂CrO₄ amount is = 331.73g* 29.3 /339.74 = 28.609 g

Thus 29.3 g of AgNO₃ is give  Ag₂CrO₄amount is = 28.609 g