Given the following reaction
Ca(OH)2+H2SO4→CaSO4+2H2O
If you start with 14.82 g of Ca(OH)2 and 16.35 g of H2SO4,
a) determine the limiting reagent
b) determine the number of moles of H2O produced
c) determine the number of grams of CaSO4 produced
d) determine the number of grams of excess reagent left
Answers
Given:
The mass of Ca(OH)2 = 14.82 gm
The mass of H2SO4 = 16.35 gm
To Find:
a) the limiting reagent
b) the number of moles of H2O produced
c) the number of grams of CaSO4 produced
d) the number of grams of excess reagent left
Calculation:
- No of moles of Ca(OH)2 reacting= 14.82/74 = 0.2
- No of moles of H2SO4 reacting = 0.167
- The chemical reaction given is:
Ca(OH)2 + H2SO4 → CaSO4 + 2 H2O
- According to the given reaction:
1 mole (98 gm) of H2SO4 reacts with 1 mole (74 gm) of Ca(OH)2 to produce 1 mole (136 gm) of CaSO4 and 2 moles (36 gm) of H2O.
⇒ 0.167 moles (16.35 gm) of H2SO4 reacts with 0.167 moles (12.36 gm) of Ca(OH)2 to produce 0.167 moles (22.71 gm) of CaSO4 and 0.334 moles (6.01 gm) of H2O.
(a) As the same amount of both the reactants is required.
- So, H2SO4 is the limiting reagent as it is reacting in less quantity than Ca(OH)2.
(b) The no of moles of H2O produced = 0.334 moles
(c) The number of grams of CaSO4 produced = 22.71 gm
(d) The number of grams of excess reagent {Ca(OH)2} left = 14.82 - 12.36 = 2.46 gm