Question

4.3120 g KNO3, 1.2345 g of CuCl2*2H2O, and 0.5678 g of BaCl2*2H2O are used to prepare K2CuBa(NO2)6. Calculate the theoretical yield of K2CuBa(NO2)6 in grams.

Answer 1

CuCl2*2H2O + BaCl2*2H2O + 6 KNO2 -----à K₂CuBa (NO₂)₆ + 4 H₂O + 4 KCl

Step 1- Calculate moles of CuCl₂*2H2O

1.2345 g of CuCl₂*2H₂O (given)

Molar mass ofCuCl₂*2H₂O = 170.48256 g/mole

Moles of CuCl2*2H2O = 1.2345 g/170.48256 g/mole = 0.00724 moles

Step 2- Calculate moles of KNO₂

4.3120 g KNO₂ (given)

Molar mass of KNO₂ = 85.1038 g /mole

Moles of KNO2 = 4.3120 g/ 85.1038 g/mole = 0.05066 moles

Step 3- Calculate moles of BaCl₂*2H₂O

0.5678 g of BaCl₂*2H₂O (given)

Molar mass of BaCl₂*₂H2O = 244.26356 g/mole

Moles of BaCl₂*2H₂O = 0.5678 g / 244.26356 g/mole = 0.00232 moles

Step 4- Find limiting reagent

Molar ratio of CuCl₂*2H₂O: BaCl₂*2H₂O: KNO₂ = 1: 1: 6

1 mole of BaCl₂*2H₂O reacts with 1 mole of CuCl₂*2H₂O

0.00232 moles of BaCl2*2H2O reacts with 0.00232 moles of CuCl₂*2H₂O

1 mole of BaCl₂*2H₂O reacts with 6 moles of KNO₂

0.00232 moles of BaCl₂*2H₂O reacts with 0.00232 x 6 = 0.01392 moles of KNO₂

So BaCl₂*2H₂O is a limiting reagent.

Step 5- Calculate moles of K₂CuBa (NO₂)₆

1 mole of BaCl₂*2H₂O produces 1 mole of K₂CuBa (NO₂)₆

So, 0.00232 moles of BaCl₂*2H₂O produce 0.00232 moles of K₂CuBa (NO₂)₆

Step 6-Calculate theoretical yield of K₂CuBa (NO₂)₆ in grams

Molar mass of K₂CuBa (NO₂)₆ = 555.1026 g/mole

Theoretical yield = moles of K₂CuBa (NO₂)₆ x molar mass of K₂CuBa (NO₂)₆

Theoretical yield = 0.00232 mole x 555.1026 g/mole = 1.2878 g