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From the equation 2C5H12OH+15O2=10CO2+12H2O

a. How many moles of oxygen are needed for the combustion of 1 mole of C5H2OH
b. How many moles of H2O are formed for each moles of O2 consumed
c. How many grams of CO2 are produced for each mole of alkane burned
d.How many grams of CO2 are produced for each gram of alkanol burned​

User Smhx
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2 Answers

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Final answer:

a. (15/2) moles of O2 are needed for the combustion of 1 mole of C5H12OH. b. For each mole of O2 consumed, 12 moles of H2O are formed. c. For each mole of C5H12OH burned, 1 mole of CO2 is produced. d. For each gram of C5H12OH burned, (1/88) moles of CO2 are produced.

Step-by-step explanation:

a. To determine the number of moles of oxygen needed for the combustion of 1 mole of C5H12OH, we can use the balanced equation:

2C5H12OH + 15O2 → 10CO2 + 12H2O

From the equation, we can see that 15 moles of O2 are needed for the combustion of 2 moles of C5H12OH. So, for the combustion of 1 mole of C5H12OH, we will need (15/2) moles of O2.

b. From the balanced equation, we can see that for each mole of O2 consumed, 12 moles of H2O are formed. So, for the combustion of 1 mole of C5H12OH, we will have 12 moles of H2O produced for each (15/2) moles of O2 consumed.

c. To determine the grams of CO2 produced for each mole of alkane burned, we can use the molar mass of CO2. The molar mass of CO2 is 44 grams/mol. From the balanced equation, we can see that for the combustion of 10 moles of C5H12OH, 10 moles of CO2 are produced. So, for each mole of C5H12OH burned, we will have (10/10) moles of CO2 produced, which is equal to 1 mole of CO2.

d. To determine the grams of CO2 produced for each gram of alkanol burned, we can use the molar mass of C5H12OH. The molar mass of C5H12OH is 88 grams/mol. From the previous calculation, we know that for each mole of C5H12OH burned, 1 mole of CO2 is produced. So, for each gram of C5H12OH burned, we will have (1/88) moles of CO2 produced.

User Davey Van Tilburg
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Answer:

a.7.5 (15/2)
b. 0.8 (12/15)
c. 220.0475g
d. 2.468g

Step-by-step explanation:

a.
1 (mole of C_5H_1_2OH) * (15 moles of O_2)/(2 moles of C_5H_1_2OH ) = 7.5 moles of O_2

b.
1 mole of O_2 * (12 moles of H_2O)/(15 moles of O_2) = (4 moles of H_2O)/(5)

c.
1 mole of C_2H_1_2OH * (10 moles of CO_2)/(2 moles of C_2H_1_2OH) * (44.0095 grams of CO_2)/(1 mole of CO_2) = 220.0475grams of CO_2

d.
1 gram of C_5H_1_2OH * (1 mole of C_5H_1_2OH)/(89.1561 grams of C_5H_1_2OH) * (10 moles of CO_2)/(2 moles of C_2H_1_2OH) * (44.0095 grams of CO_2)/(1 mole of CO_2) = 220.0475grams of CO_2

User Bsferreira
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