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

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.

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