Based on the following balanced equation:

2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O

a. How many moles of CO2 are produced after complete reaction of 8.00 moles of C4H10?
b. How many grams of CO2 are produced after complete reaction of 8.00 moles of C4H10?
c. How many moles of H2O are produced after complete reaction of 8.00 moles of C4H10?
d. How many grams of H2O are produced after complete reaction of 8.00 moles of C4H10?

My responses:
a. 32.0 mol CO2
b. 1.41 x 10^3 g CO2
c. 40.0 mol H2O
d. 720 g H2O

To find the answers to these questions, we can use the given balanced equation and the concept of mole ratios.

a. To find the number of moles of CO2 produced, we first look at the balanced equation: 2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O

According to the equation, 2 moles of C4H10 react to produce 8 moles of CO2. Therefore, if we start with 8.00 moles of C4H10, we can use the ratio:

8.00 mol C4H10 * (8 mol CO2 / 2 mol C4H10) = 32.0 mol CO2

So, the answer is 32.0 mol CO2.

b. To find the grams of CO2 produced, we need to know the molar mass of CO2, which is 44.01 g/mol. Using this information, we can convert the moles of CO2 to grams:

32.0 mol CO2 * (44.01 g CO2 / 1 mol CO2) = 1.41 x 10^3 g CO2

So, the answer is 1.41 x 10^3 g CO2.

c. Similar to part (a), we can use the balanced equation to determine the number of moles of H2O produced. According to the equation, 2 moles of C4H10 react to produce 10 moles of H2O. Therefore, using the same ratio as before:

8.00 mol C4H10 * (10 mol H2O / 2 mol C4H10) = 40.0 mol H2O

So, the answer is 40.0 mol H2O.

d. To find the grams of H2O produced, we need to know the molar mass of H2O, which is 18.02 g/mol. Using this information, we can convert the moles of H2O to grams:

40.0 mol H2O * (18.02 g H2O / 1 mol H2O) = 720 g H2O

So, the answer is 720 g H2O.

By using the balanced equation and the concept of mole ratios, we can calculate the number of moles and grams of CO2 and H2O produced when 8.00 moles of C4H10 react.

a. To find the number of moles of CO2 produced, we can use the stoichiometric ratio between C4H10 and CO2. According to the balanced equation, 2 moles of C4H10 react to produce 8 moles of CO2. Therefore, for every 2 moles of C4H10, we have 8 moles of CO2.

Since we have 8.00 moles of C4H10, we can set up the following ratio:

2 moles C4H10 : 8 moles CO2
8.00 moles C4H10 : x moles CO2

Using the ratio, we can calculate the number of moles of CO2:

(8.00 moles C4H10) x (8 moles CO2 / 2 moles C4H10) = 32.0 moles CO2

Therefore, 32.0 moles of CO2 are produced after the complete reaction of 8.00 moles of C4H10.

b. To find the mass of CO2 produced, we need to use the molar mass of CO2. The molar mass of CO2 is the sum of the atomic masses of carbon (C) and oxygen (O).

Molar mass of CO2 = (atomic mass of C) + 2 x (atomic mass of O)
= (12.01 g/mol) + 2 x (16.00 g/mol)
= 44.01 g/mol

Now, we can calculate the mass of CO2 produced:

Mass of CO2 = moles of CO2 x molar mass of CO2
= 32.0 moles CO2 x 44.01 g/mol
≈ 1.41 x 10^3 g CO2

Therefore, approximately 1.41 x 10^3 grams of CO2 are produced after the complete reaction of 8.00 moles of C4H10.

c. Similarly, to find the number of moles of H2O produced, we can use the stoichiometric ratio between C4H10 and H2O. According to the balanced equation, 2 moles of C4H10 react to produce 10 moles of H2O. Therefore, for every 2 moles of C4H10, we have 10 moles of H2O.

Using this ratio, we can calculate the number of moles of H2O:

(8.00 moles C4H10) x (10 moles H2O / 2 moles C4H10) = 40.0 moles H2O

Therefore, 40.0 moles of H2O are produced after the complete reaction of 8.00 moles of C4H10.

d. Finally, to find the mass of H2O produced, we need to use the molar mass of H2O. The molar mass of H2O is the sum of the atomic masses of hydrogen (H) and oxygen (O).

Molar mass of H2O = 2 x (atomic mass of H) + atomic mass of O
= 2 x (1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol

Now, we can calculate the mass of H2O produced:

Mass of H2O = moles of H2O x molar mass of H2O
= 40.0 moles H2O x 18.02 g/mol
= 720 g H2O

Therefore, 720 grams of H2O are produced after the complete reaction of 8.00 moles of C4H10.

Very good. Go to the top of the class.