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There are 237. 5 g of Sulfur,S in 475 g of SO2?
<h3 /><h3>Calculation of grams of Sulfur</h3>From the question, we can say that
- The molar mass of sulfur = 32 g/mol
- The molar mass of oxygen = 16 g/mol
Therefore,
The molar mass for SO2 = 32 + (16 × 2) g/mol = 64 g/mol
Now,
If 1 mole of SO2 contains 1 mole of S
Then 64 g of SO2, will contain 32g of S;
Such that
475 g of SO2 will give { } = 237. 5 g of Sulfur.
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Answer:
CaO is the limiting reagent
Theoritical yield = 25.71 g
% Yield = 75.44%
Explanation:
1 mole = Molar mass of the substance
Molar Mass of CaO = 56 g/mol
Molar Mass of CaCO3 = 100 g/mol
Molar mass of CO2 = 44 g/mol
The balanced Equation is :
1 mole of CaO reacts with = 1 mole of CO2
56 g of CaO reacts with = 44 g of CO2
1 g of CaO reacts with =
= 0.785 g of CO2
So,
<u>14.4 g of CaO</u><u> </u>must react with = (14.4 x 0.785) g of CO2
= 11.31 g of CO2
<u>Needed = 11.31 g</u>
<u>Available CO2 = 13.8 g</u><u> </u>(given)
So CO2 is in excess , hence<u> CaO is the limiting reagent and product will produce from 14.4 g of CaO</u>
1 mole of CaO will produce 1 mole pf CaCO3
56 g of CaO produce = 100 g of CaCO3
1 g of CaO produce =
= 1.785 g of CaCO3
14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3
= 25.71 g of CaCO3
Theoritical Yield of CaCO3 = 25.71 g
Actual yield = 19.4 g
Percent Yield =
= 75.44 %