CHEMISTRY| MOLE CONCEPT | Question 05
Determine the limiting reagent of this reaction.
Na2B4O7 + H2SO4 + 5H2O → 4H3BO3 + Na2SO4
There are 5.00g of each reactant.
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DATA GIVEN:
Mass of each reagent was 5.00g
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Limiting reagent
Mole for Na2
= mass /Mole mass
= 5g /186g/mole
= 0.02688 mole
Mole ratio of Na2 = 0.002688 /1
= 0.02688
Mole for H2SO4 = mass /Molar mass
= 5g /98g/mole
= 0.051 mole
Mole ratio for H4SO4 = 0.051 mole /1 mole
= 0.05
Mole for H2O = mass /Molar mass
= 5g /18g/mole
= 0.277 mole
Mole ratio for H2 = 0.277 /5
= 0.0556
The limiting reagent is Na2B4O7
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