A. Which reagent is the limiting reactant when 0.420 mol Al(OH)3 and 0.420 mol H2SO4 are enabled to react?

1. ) H2SO4

2. ) Al(OH)3

3. ) Al2(SO4)3

4. ) H2O

B. How many moles of Al2(SO4)3 deserve to develop under these conditions? answer in mol

How many kind of moles of the excess reactant remajor after the completion of the reaction? answer in mol

Aluminum hydroxide reacts with sulfuric acid as follows:

2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l).

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The coefficients in well balanced equation suggest the ratio of moles of assets to moles of reactants.

2 moles of Al(OH)3 react through 3 moles of H2SO4 to develop 1 mole of Al2(SO4)3 and also 6 moles of H2O.

A. Which reagent is the limiting reactant once 0.420 mol Al(OH)3 and 0.420 mol H2SO4 are enabled to react?

The limiting reactant is the reactant which you execute not have actually enough of!

Ratio of Al(OH)3 : H2SO4 = 2 : 3

So, 0.420 mole of Al(OH)3 calls for 3/2 * 0.420 mole of H2SO4 = 0.630 mole of H2SO4. You only have actually 0.420 mole of H2SO4, so H2SO4 is the limiting reactant.

And

0.420 mole of H2SO4 needs ⅔ * 0.420 mole of Al(OH)3 = 0.280 mole of Al(OH)3. You have actually excess moles of Al(OH)3, so Al(OH)3 is in excess

Excess 0.420 – 0.280 =

Ratio of H2SO4 : Al2(SO4)3 = 3 : 1

0.420 mole of H2SO4 will certainly produce ⅓ * 0.420 = 0.14 mole of Al2(SO4)3

A. Since you are starting via equal moles of the two reactants, you have the right to view from the well balanced equation that you require 3 moles of H2SO4 for eincredibly 2 moles of Al(OH)3. So, H2SO4 is the limiting reactant.

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B. 0.420 mol H2SO4 X (1 mol Al2(SO4)3 / 3 mol H2SO4) = 0.140 mol Al2(SO4)3

C. Moles of Al(OH)3 used:

0.420 mol H2SO4 X (2 mol Al(OH)3/3 mol H2SO4) = 0.280 mol Al(OH)3 used

Moles continuing to be = 0.420 - 0.280 = 0.140 mol Al(OH)3 remaining

0.223 mols of Aluminium Sulphate have the right to form

0.223mols

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