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19 March, 03:35

Hydrogen gas and oxygen gas can be combined to produce liquid water. If 15.0 g of each gas is combined,

a. W hat mass of water will be produced, assuming the reaction runs to completion?

b. Which reactant will remain, and how many grams?

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  1. 19 March, 06:56
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    a) There will be produced 16.89 grams of H2O

    b) There will remain 6.4885 moles of H2. This is 13.1 grams H2

    Explanation:

    Step 1: Data given

    Mass of hydrogen gas = 15.0 grams

    Mass of oxygen gass = 15.0 grams

    Molar mass of hydrgen = 2.02 g/mol

    Molar mass of oxygen = 32 g/mol

    Step 2: The balanced equation:

    2H2 + O2 → 2H2O

    Step 3: Calculate moles of H2

    Moles H2 = Mass H2 / Molar mass H2

    Moles H2 = 15.00 g / 2.02 g/mol

    Moles H2 = 7.426 moles

    Step 4: Calculate moles of O2

    Moles O2 = 15.00 g / 32g/mol

    Moles O2 = 0.46875 moles

    Step 5: Calculate limiting reactant

    For 2 moles of H2 we need 1 mol of O2 to produce 2 moles of H2O

    O2 is the limiting reactant. It will completely be consumed (0.46875 moles).

    H2 is in excess. There will be consumed 0.46875 * 2 = 0.9375 moles of H2

    There will remain 6.4885 moles of H2. This is 13.1 grams

    Step 6: Calculate moles of H2O

    For 2 moles of H2 we need 1 mol of O2 to produce 2 moles of H2O

    For 0.46875 moles we will have 0.9375 moles of H2O produced.

    Step 7: Calculate mass of H2O

    Mass H2O = moles H2O * molar mass H2O

    Mass H2O = 0.9375 moles * 18.02 g/mol

    Mass H2O = 16.89 grams

    There will be produced 16.89 grams of H2O
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