C4H10 + 6.5O2 → 4CO2 + 5H2O?

Calculate the number of grams of H2O produced when 7g of C4H10 is burnt in oxygen

Please can you explain in simple terms how you worked this out

Thank you so much

Comments

  • Look at the balanced equation: This tells you that 1 mol of C4H10 will produce 5 mol H2O

    How many mol of C4H10 do you have in 7g ?

    Molar mass C4H10 = 4*12 + 10*1 = 58g/mol

    Moles in 7g = 7/58 = 0.121 mol C4H10

    This will produce 0.121*5 = 0.605 mol H2O

    Mo9lar mass of H2O = 18g/mol

    Mass of 0.605 mol H2O = 0.605*18 = 10.89g H2O produced

    You can have only 1 significant figure in the answer - 10g H2O produced.

  • Basically, convert the amount of grams of C4H10 to moles, compare the mole ratio between C4H10 and H2O, and convert the amount of moles of H2O to grams.

    1mol C4H10 = 4*12.0107g + 10*1.00794g = 58.1222g C4H10

    7g of C4H10 * 1mol of C4H10/58.1222g of C4H10 = 0.1204359092mol of C4H10

    For every mole of C4H10 reacting should produce 5 moles of H2O; therefore, the ratio between moles of H2O and C4H10 is 5 to 1.

    0.1204359092mol of C4H10 * 5mol of H2O/1mol of C4H10 = 0.6021795459mol of H2O

    1mol H2O = 2*1.00794g + 15.9994g = 18.01528g H2O

    0.6021795459mol of H2O * 18.01528g of H2O/1mol of H2O = 10.84843313g of H2O

    Hope this helps

  • Equation: C4H10 + 6.5O2 → 4CO2 + 5H2O

    moles of:......1..........6.5............4.............5

    mass shown:58 g....6.5x16 g...4x44g.....5x18 g

    .......................58 g 208 g......176 g ......108 g

    ? g H2O = 7 g C4H10 x 108 g H2O / 58 g C4H10 = 13 g H2O

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