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Today, 08:51

Carbon disulfide burns in oxygen to yield carbon dioxide and sulfur dioxide according to the following chemical equation. CS2 (l) + 3O2 (g) → CO2 (g) + 2SO2 (g)

a. If 1.00 mol CS2 reacts with 1.00 mol O2, identify the limiting reactant.

b. How many moles of excess reactant remain?

c. How many moles of each product are formed?

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  1. Today, 09:30
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    a) the limiting reactant is 02

    b) There will remain 0.667 moles of CS2

    c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2

    Explanation:

    Step 1: Data given

    Number of moles of CS2 = 1.00 mol

    Number of moles of O2 = 1.00 mol

    Molar mass of O2 = 32 g/mol

    Molar mass of CS2 = 76.14 g/mol

    Step 2: The balanced equation

    CS2 (l) + 3O2 (g) → CO2 (g) + 2SO2 (g)

    Step 3: Calculate the limiting reactant

    For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

    O2 is the limiting reactant. It will completely be consumed. (1.00 mol).

    CS2 is in excess. There will react 1.00 / 3 = 0.333 moles

    There will remain 1.00 - 0.333 = 0.667 moles of CS2

    Step 4: Calculate moles of CO2 and SO2

    For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

    For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2

    For 1.00 mol of O2 we have 1.00 / (3/2) = 0.667 moles of SO2
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