Please slove 25-28 and show all the steps For questions 25-28, consider the comb

Question

Please slove 25-28 and show all the steps

For questions 25-28, consider the combustion in 85.9 g of propane. C_3H_5 with excess oxygen on, O_3, to produce CO_2 and H_2O. Write the 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? A) 1 B) 3 C) 5 D) 7 How many moles of propane were used? A) 85.9 B) 12.01 C) 1.95 D) 084 How many moles of CO_2 were produced? A) 7.80 B) 0.25 C) 0.65 D) 584 How many grams of CO_2 were produced? A) 195 B) 257 C) 22.4 D) 106

Explanation / Answer

The balanced combustion reaction of propane is :

C3H8 + 5O2 = 3CO2 + 4H2O

Ans . 25 . C) 5

5 moles of O2 react with 1 mole of propane to form 3 moles of CO2 and 4 moles of H2O.

Ans 26. C) 1.95

The molar mass of propane is 44.1 g / mol

So 85.9 g = 85.9 / 44.1 = 1.95 moles

Ans 27 . D) 5.84

1 mole of propane makes 3 moles of CO2

so 1.95 moles will make 1.95 x 3 = 5.84 moles of CO2

Ans 28.

Molar mass of CO2 = 44 g/ mol

5.84 x 44 = 257 g

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