chemistry (quickly please)

Question

Write a balanced equation for the reaction of sulfuric acid

(H2(SO4)) with gallium (Il) hydroxide to form water and gallium (lI) sulfate

1. From the balanced equation, determine the mass of gallium sulfate that can be made with 145 grams of sulfuric acid and 320 grams of gallium hydroxide.
2. Which is the limiting reagent?
3. How much of the excess reagent will remain once the reaction?

If the actual yield of Ga(SO4) in a lab experiment is 237 grams, what is the percent yield?

Answer 1

Note: Gallium(II) hydroxide is not a common compound, as the +3 oxidation state of Ga is much more common. If you meant to use gallium(III) hydroxide, please re-submit the question.

Sulfuric acid: H₂SO₄ (molar mass, 98.08 g/mol)

Gallium(II) hydroxide: Ga(OH)₂ (molar mass, 103.7 g/mol)

Gallium(II) sulfate: GaSO4 (molar mass, 165.8 g/mol)

H₂SO₄ + Ga(OH)₂ ==> GaSO₄ + 2H₂O .. balanced equation

Before one can calculate part (1), one must determine the answer to part (2) .. which is the limiting reactant? One way to determine this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less will represent the limiting reagent.

For H₂SO₄: 145 g H₂SO₄ x 1 mol / 98.08 g = 1.478 moles (÷1->1.48)

For Ga(OH)₂: 320 g x 1 mol / 103.7 g = 3.086 moles (÷1->3.09)

Part (2): Since 1.48 is the lesser value, this tells us that H₂SO₄ is the limiting reagent and we must use 1.478 moles to determine the amount of product (GaSO₄) that can be formed.

Part (1): Mass of GaSO₄ = 1.478 mols H₂SO₄ x 1 mol GaSO₄ / mol H₂SO₄ x 165.8 g/mol = 245 g GaSO₄