Test: Stoichiometry and Stoichiometric Calculations (April 8) - JEE MCQ

2 g of H₂ reacts with 71 g of CI₂
100 g of H₂ will react with 71/2 x 100 = 3550g of CI₂
Hence, CI₂ is the limiting reagent.
71 g of CI₂ produces 73 g of HCl
100g of CI₂ will produce 73/71 x 100 = 102.8 g of HCl

100 g of CaCO₃ reacts with 73 g of HCl
40 g of CaCO₃ will react with 73/100 x 40 = 29.2 g of HCl
Since CaCO₃ is completely consumed and some amount (40 - 29.2 = 10.8g) of HCl remains unreacted and hence, CaCO₃ is limiting reagent.

28 g of C₂H₄ requires 96 g of O₂
560 g of C₂H₄ requires 96/28 x 560
= 1920 g or 1.92 kg of O₂

Molar mass of sodium acetate (CH₃COONa)
= 82.0245 g/mol
Mass of CH₃COONa required to make 250 mL of 0.575 M solution 

Mass percent of X

No. of moles of glucose = 2.82/180 = 0.01567
No. of moles of water = 30/18 = 1.667
Total no. of moles of solution = 0.01567 + 1.667 = 1.683
Mole fraction of glucose = 0.01567/1.683 = 0.0093 = 0.01

M₁V₁ = M₂V₂ = MV
0.5 x 50 + 0.25 x 150 = M x 250
M = (25 + 37.5)/250 = 0.25 M

Molar mass of CUSO₄ = 63.5 + 32 + 64 = 159.5
Moles of CUSO₄ = 80/159.5 = O.50
Volume of solution = 3 L

= 0.167 mol L^-1

Na₂SO₄ + BaCI₂ → BaSO₄ + 2NaCl
No. of moles of BaSO₄ = w/M = 10/233 = 0.0429
∴ No. of moles of Na₂SO₄ needed = M x V/1000
Or 0.0429 = 5 x V/1000
V = 8.58 mL

Molality =