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Isooctane is a fuel additive that prevents premature and spontaneous ignition under the heat and pressure...

23. Isooctane is a fuel additive that prevents premature and spontaneous ignition under the heat and pressure of an engine's cylinder. The combustion of isooctane is shown below:
2 C8H18(l) + 25 O2(g) -> 16 CO2(g) + 18 H2O(g)
How many grams of water (H2O, MM = 18.015 g/mol) are produced when 393 g of isooctane (C8H18, MM = 114.23 g/mol) reacts with excess oxygen?
24. What volume, in milliliters, of 0.78 M H3PO4(aq) is needed to completely react with 637 mL of 0.58 M Mg(OH)2(aq) solution?
2 H3PO4(aq) + 3 Mg(OH)2(aq) -> 6 H2O(l) + Mg3(PO4)2(aq)
A. 473 mL H3PO4
B. 425 mL H3PO4
C. 316 mL H3PO4
D. 711 mL H3PO4
25. Historians believe that ancient people produced copper metal from cuprite (Cu2O) according to the balanced chemical reaction below:
2 Cu2O(s) + C(s) -> 4 Cu(s) + CO2(g)
If a mixture contains 1000. grams of Cu2O (145.00 g/mol) and 445.0 grams of C (12.011 g/mol), what is the limiting reactant?
A. Cu2O
B. C
C. Cu
D. CO2

Answer

  1. 23.

    To find the grams of water produced, first calculate the moles of isooctane:
    Moles of C8H18 = 393 g / 114.23 g/mol = 3.44 mol
    From the balanced equation, 2 moles of C8H18 produce 18 moles of H2O. Therefore, 3.44 moles of C8H18 produce:
    (18/2) * 3.44 = 30.96 moles of H2O
    Grams of H2O = 30.96 moles * 18.015 g/mol = 558 g
    Answer: 558 g H2O

  2. 24.

    Using the stoichiometry of the reaction:
    Moles of Mg(OH)2 = 0.637 L * 0.58 mol/L = 0.36946 mol
    From the balanced equation, 3 moles of Mg(OH)2 react with 2 moles of H3PO4. Therefore, moles of H3PO4 needed:
    (2/3) * 0.36946 = 0.2463 mol
    Volume of H3PO4 = 0.2463 mol / 0.78 mol/L = 0.315 L = 315 mL
    Answer: C. 316 mL H3PO4

  3. 25.

    Calculate moles of each reactant:
    Moles of Cu2O = 1000 g / 145.00 g/mol = 6.897 mol
    Moles of C = 445.0 g / 12.011 g/mol = 37.049 mol
    From the balanced equation, 2 moles of Cu2O react with 1 mole of C. Therefore, moles of C needed for 6.897 moles of Cu2O:
    6.897 / 2 = 3.4485 mol
    Since 3.4485 mol of C is needed and 37.049 mol is available, Cu2O is the limiting reactant.
    Answer: A. Cu2O

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