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22 July, 12:11

For each of the following unbalanced reactions, suppose 5.17 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent (s) will remain after the limiting reactant is consumed. Na2B4O7 (s) + H2SO4 (aq) + H2O (l) → H3BO3 (s) + Na2SO4 (aq)

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  1. 22 July, 13:35
    0
    Limiting reactant is the Na₂B₄O₇

    Mass of H₂SO₄ that remains after the reaction: 2.65 g

    Mass of water that remains after the reaction: 2.86 g

    Explanation:

    We analyse the reaction:

    Na₂B₄O₇ (s) + H₂SO₄ (aq) + 5H₂O (l) → 4H₃BO₃ (s) + Na₂SO₄ (aq)

    So we have 5.17g of borax, 5.17 g of sulfuric acid and 5.17 g of water.

    We convert the mass to moles:

    5.17g / 201.24 g/mol = 0.0257 moles of borax

    5.17 g / 98 g/mol = 0.0527 moles of H₂SO₄

    5.17 g / 18 g/mol = 0.287 moles of H₂O

    Ratio between borax and sulfuric acid is 1:1. For 0.0257 moles of borax I need the same amount of acid, so the acid is in excess. For 0.0527 moles of acid I need the same amount of borax, but I do not have enough moles, so the borax is the limiting reactant.

    1 mol of borax need 5 moles of water to react

    Then, 0.0257 moles of borax will react with (0.0257. 5) / 1 = 0.128 moles of water. Water is also an excess because I have enough moles to use.

    5 moles of water react with 1 mol of borax

    Therefore 0.287 moles of water will react with (0.287. 1) / 5 = 0.0574-

    There is no enough borax to react. It's ok that Na₂B₄O₇ is the limiting reactant so:

    1 mol of borax react with 1 mol of acid and 5 moles of water

    Then, 0.0257 moles of borax will react with 0.0257 moles of acid and 0.128 moles of H₂O

    Moles of sulfuric acid that remains: 0.0527 - 0.0257 = 0.027 moles

    Moles of water that remains: 0.287 - 0.128 = 0.159 moles

    Mass of H₂SO₄: 0.027 mol. 98g / 1mol = 2.65 g

    Mass of water: 0.159 mol. 18 g / 1 mol = 2.86 g
  2. 22 July, 13:53
    0
    Na2B4O7 is the limiting reactant. There will remain 2.65 grams of H2SO4 and 2.86 grams of H2O

    Explanation:

    Step 1: Data given

    Mass of Na2B4O7 = 5.17 grams

    Mass H2SO4 = 5.17 grams

    Mass H2O = 5.17 grams

    Molar mass of Na2B4O7 = 201.22 g/mol

    Molar mass H2SO4 = 98.08 g/mol

    Molar mass H2O = 18.02 g/mol

    Step 2: The balanced equation

    Na2B4O7 (s) + H2SO4 (aq) + 5H2O (l) → 4H3BO3 (s) + Na2SO4 (aq)

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles Na2B4O7 = 5.17 grams / 201.22 g/mol

    Moles Na2B4O7 = 0.0257 moles

    Moles H2SO4 = 5.17 grams / 98.08 g/mol

    Moles H2SO4 = 0.0527

    Moles H2O = 5.17 grams / 18.02 g/mol

    Moles H2O = 0.287 moles

    Step 3: Calculate the limiting reactant

    For 1 mol Na2B4O7 we need 1 mol H2SO4 and 5 moles H2O to produce 4 moles H3BO3 and 1 mol Na2SO4

    Na2B4O7 is the limiting reactant. It will completely be consumed (0.0257 moles). H2SO4 and H2O are in exces. There will react 0.0257 moles H2SO4 and 5*0.0257 = 0.1285 moles H2O

    There will remain 0.0527 - 0.0257 = 0.027 moles H2SO4

    There will remain 0.287 - 0.1285 = 0.1585 moles H2O

    Step 4: Calculate mass of excess reactants

    Mass = moles * molar mass

    Mass H2SO4 remaining = 0.027 moles * 98.08 g/mol

    Mass H2SO4 = 2.65 grams

    Mass H2O remaining = 0.1585 moles * 18.02 g/mol

    Mass H2O remaining = 2.86 grams

    Na2B4O7 is the limiting reactant. There will remain 2.65 grams of H2SO4 and 2.86 grams of H2O
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