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)

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

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

fe 55.85 g/mol and h2o 18.02 g/mol.