Question

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4 FeS₂   +   11 O₂    ®    2 Fe₂O₃    +    8 SO₂

1.    How many moles of oxygen are required to react with 3.75 moles of FeS₂?

      

2.    How many moles of Fe₂O₃ are formed when 2.83 moles of oxygen reacts with

       excess FeS₂?

3.    How many moles of SO₂ are formed when 2.81 moles of Fe₂O₃ are formed?

4.    How many grams of SO₂are formed when 1.00 mol of FeS₂ reacts?

5.    How many moles of O₂ are required to from 75.8 grams of Fe₂O₃?

6.    How many grams of Fe₂O₃ are formed when 187 grams of FeS₂ react with excess   

       oxygen?

7.    How many grams of oxygen are required to form 67.8 grams of Fe₂O₃?

8.    How many grams of FeS₂ are required to react with excess oxygen to produce

       77.0 grams of SO₂?

9.    How many molecules of SO₂ are formed when 27.0 grams of FeS₂ react?

Percent Yield

10.   If 30.7 grams of Fe₂O₃ are obtained from the reaction of 85.3 grams of FeS₂ with

       excess oxygen, what is the percent yield?

11.   What is the percent yield if 130 grams of SO₂ are obtained from the reaction of        140 grams of FeS₂ with excess oxygen?

Limiting Reactant

12.   What is the maximum amount of Fe₂O₃ that can be prepared from 57.8 grams of

       FeS₂ and 92.5 grams of O₂? What is the limiting reactant?

13.   How many grams of SO₂ can be prepared from 86.2 grams of FeS₂ and

       125.3 grams of oxygen? How many grams of each reactant and product are     present at the end of the reaction?

14.   What is the percent yield if 118 grams of SO₂ are produced from the reaction of

       135 grams of FeS₂ and 200 grams of oxygen?

Answer 1

moles = grams/molar mass

In the above reaction, 4 mols of FeS2 reacts with 11 mols of O2 to form 2 mols of Fe2O3 and 8 mols of SO2

1) moles of O2 required to react with 3.75 mols of FeS2 = 3.75 x 11/4 = 10.31 mols

2) moles of Fe2O3 formed with 2.83 mols of O2 = 2.83 x 2/11 = 0.51 mols

3) moles of SO2 formed when 2.81 mols of Fe2O3 is formed = 2.81 x 8/2 = 11.24 mols

4) grams of SO2 formed when 1 mol of FeS2 reacts = 8 x 64.066/4 = 128.13 g

5) moles of O2 required to form 75.8 g of Fe2O3

moles of Fe2O3 = 75.8/159.69 = 0.47 mols

mols of O2 required = 0.47 x 11/2 = 2.61 mols

6) moles of FeS2 reacted = 187/119.98 = 1.56 mols

grams of Fe2O3 formed = 1.56 x 2 x 159.69/4 = 124.44 g

7) moles of Fe2O3 formed = 67.8/159.69 = 0.42 mols

moles of O2 required = 0.42 x 11/2 = 2.33 mols

grams of O2 required = 2.33 x 31.9988 = 74.72 g

8) moles of SO2 formed = 77/64.066 = 1.20 mols

grams of FeS2 required = 1.20 x 4 x 119.98/8 = 72.10 g

9) moles of FeS2 reacted = 27/119.98 = 0.225 mols

moleules of SO2 formed = (0.225 x 8/4) mols x 6.022 x 10^23 = 2.71 x 10^23 molecules.

Percent yield

10) theoretical yield of Fe2O3 from 85.3 g of FeS2

= 2/4 x (85.3/119.98) x 159.69 = 56.69 g

% yield = (30.7/56.69) x 100 = 54.15%

11) moles of FeS2 = 140/119.98 =1.17 mols

theoretical mass of SO2 formed = 1.17 x 8 x 64.066/4 = 149.51 g

% yield = (130/149.51) x 100 = 86.95%

Limiting reagent

12) moles of FeS2 = 57.8/119.98 = 0.482 mols

moles of O2 = 92.5/31.9988 = 2.891 mols

If all of FeS2 is consumed we would require = 11 x 0.482/4 = 1.32 moles of O2

If all of O2 is consumed we would require = 4 x 2.891/11 = 1.05 moles of FeS2

Since the moles of O2 is in excess to required moles, FeS2 is the limiting reagent

maximum amount of Fe2O3 formed = 0.482 x 2 x 159.69/4 = 38.48 g

13) moles of FeS2 = 86.2/119.98 = 0.72 mols

moles of O2 = 125.3/31.9988 = 3.91 mols

If all of FeS2 is consumed we would require = 1.98 mols of O2

If all of O2 is consumed we would require = 1.42 mols of FeS2

Since moles of FeS2 is less then required, this is the limiting reagent.

grams of FeS2 present at the end = 0 g

grams of O2 present at the end = (3.91-0.72) x 31.9988 = 102.08 g

grams of Fe2O3 present at the end = 0.72 x 2 x 159.69/4 = 57.49 g

grams of SO2 prepared = 0.72 x 8 x 64.066/4 = 92.25 g

14) moles of SO2 formed = 118/64.066 = 1.842 mols

moles of FeS2 = 135/119.98 = 1.125 mols

moles of O2 = 200/31.9988 = 6.250 mols

If all of FeS2 is consumed we would require = 11 x 1.125/4 = 3.094 mols of O2

If all of O2 is consumed we would require = 4 x 6.250/11 = 2.273 mols of FeS2

Theoretical yield of SO2 = 64.066 = 1.125 x 8 x 64.066/4 = 144.15 g

% yield = (118/144.15) x 100 = 81.86%