Please slove 29 to 32 clearly and full steps. Thanks For questions 29-32, consid

Question

Please slove 29 to 32 clearly and full steps. Thanks

For questions 29-32, consider the combustion in 85.9 g of propane, C_3 H_8 with excess oxygen. O_2, to produce CO_2, and H_2 O. Write and balance the reaction. the find the mass of CO_2 produced. Then, answer the following questions. What is the coefficient on O_2 in the balanced reaction? 1 3 5 7 How many moles of propane were used? 85.9 12.01 1.95 0.84 How many moles of CO_2 were produced? 7.80 0.25 0.65 5.84 How many grams of CO_2 were produced? 195 257 22.4 106

Explanation / Answer

The chemical reaction taking place is the combustion of propane, C3H8. C3H8 burns completely to produce carbon dioxide, CO2 and water, H2O.

29. The chemical equation taking place is

C3H8 + O2 ------> CO2 + H2O

Since C3H8 is the only source of carbon and hydrogen, hence we need to balance the number of C and H atoms on both sides. Multiply CO2 by 3 and H2O by 4 to balance C and H atoms on both sides. This gives

C3H8 + O2 ------> 3 CO2 + 4 H2O

We have (3*2 + 4*1) = 10 O atoms on the right and only 2 on the left. Hence, multiply O2 by 5 to balance O atoms on both sides. This gives us the balanced chemical equation as

C3H8 + 5 O2 ------> 3 CO2 + 4 H2O …….(1)

Ans: (C)

30. We have 85.9 g of C3H8.

Molar mass of C3H8 = (3*12.01 + 8*1.008) g/mol = 44.094 g/mol.

Moles of C3H8 = (mass of C3H8 taken)/(molar mass of C3H8) = (85.9 g)/(44.094 g/mol) = 1.9481 mole 1.95 mole.

Ans: (C)

31. As per the balanced stoichiometric equation (1), we have

1 mole C3H8 = 3 moles CO2

Therefore, 1.95 mole C3H8 = (1.95 mole C3H8)*(3 moles CO2/1 mole C3H8) = 5.85 moles CO2.

Ans: The closest option is (D), hence (D) is the correct option.

32. Molar mass of CO2 = (1*12.01 + 2*15.9994) g/mol = 44.0088 g/mol.

Mass of CO2 produced = (moles of CO2)*(molar mass of CO2) = (5.84 mole)*(44.0088 g/mol) = 257.011 g 257 g.

Ans: (B)

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