Exams › JEE Main › Chemistry

When carbon dioxide gas is passed over red-hot coke, part of it is converted into carbon monoxide. If 0.5 L of CO₂ is passed over red-hot coke and the total volume of the resulting gaseous mixture becomes 700 mL, what is the composition of the gas mixture at STP?

1. CO₂ = 300 mL; CO = 400 mL
2. CO₂ = 400 mL; CO = 700 mL
3. CO₂ = 200 mL; CO = 500 mL
4. CO₂ = 350 mL; CO = 350 mL

Correct answer: CO₂ = 300 mL; CO = 400 mL

Solution

If x mL CO2 reacts, total gas = (500 - x) + 2x = 500 + x = 700, so x = 200. Remaining CO2 = 300 mL and CO = 400 mL, not 200/500.

Related JEE Main Chemistry questions

• A 100 mL sample of a urea solution contains 6.02 × 10²⁰ molecules of urea. What is the molarity of the solution? (Take Avogadro constant, N_A = 6.02 × 10²³ mol^−1.)
• When 4.9 g of H2SO4 reacts with NaCl, the products formed are 6 g of sodium hydrogen sulphate and 1.825 g of HCl. What mass of NaCl was decomposed?
• Which of the following has the least mass?
• What amount of magnesium phosphate, Mg3(PO4)2, in moles is needed to provide 0.25 mole of oxygen atoms?
• Haemoglobin has 0.33% iron by mass. If its molecular mass is about 67200 and the atomic mass of iron is 56, how many iron atoms are present in a single haemoglobin molecule?
• For the reaction 2Al(s) + 6HCl(aq) → 2Al³+(aq) + 6Cl^−(aq) + 3H₂(g), which statement is correct?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →