PYQ Vault

Combustion stoichiometry of a fuel: mass or moles of fuel, oxygen, CO2 or water

This type of question, asked 7 times: 2021 Paper 14, 14 Jun 2022, 26 Jun 2022, 27 Jan 2024, 31 Jan 2024, 5 Apr 2024, 8 Apr 2024. All 7 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 14

    27
    In the complete combustion of Butane, 72 g72\text{ }g of water is given out. How much Butane was there?

    Tap an option to check your answer.

  2. Question 2

    14 Jun 2022

    31
    If a rocket runs on a fuel (C15H30)\left( C_{15}H_{30} \right) and liquid oxygen, the weight of oxygen required and CO2CO_{2} released for every litre of fuel respectively are: (Given: density of the fuel is 0.756 g/mL0.756\text{ }g/mL )

    Tap an option to check your answer.

  3. Question 3

    26 Jun 2022

    150
    The moles of methane required to produce 81 g81\text{ }g of water after complete combustion is ×10−2\times10^{- 2} mol. [nearest integer]
  4. Question 4

    27 Jan 2024

    53
    Mass of methane required to produce 22 g22\text{ }g of CO2CO_{2} after complete combustion is ____ g. (Given Molar mass in g mol-1 C=12.0C = 12.0
    H=1.0O=16.0)\begin{matrix} & H = 1.0 \\ & O = 16.0) \end{matrix}
  5. Question 5

    31 Jan 2024

    53
    Number of moles of methane required to produce 22 gCO2( g)22\text{ }gCO_{2(\text{ }g)} after combustion is x×10−2x \times10^{- 2} moles. The value of xx is
  6. Question 6

    5 Apr 2024

    135
    The number of moles of methane required to produce 11 gCO2( g)11\text{ }gCO_{2}(\text{ }g) after complete combustion is: (Given molar mass of methane in mol−1:16mol^{- 1}:16 )

    Tap an option to check your answer.

  7. Question 7

    8 Apr 2024

    33
    Combustion of glucose (C6H12O6)\left( C_{6}H_{12}O_{6} \right) produces CO2CO_{2} and water. The amount of oxygen (in gg ) required for the complete combustion of 900 g900\text{ }g of glucose is: [Molar mass of glucose in mol−1=180mol^{- 1}= 180 ]

    Tap an option to check your answer.