How many grams C3H7OH can be made by reacting with 7.3L of CO2 at STP

6.54g of C3H7OH

Explanation:

Step 1:

Determination of the number of mole of CO2 that occupy 7.3L at stp.

This can be obtained as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, Xmol of CO2 will occupy 7.3L at stp i.e

Xmol of CO2 = 7.3/22.4

Xmol of CO2 = 0.326 mole.

Therefore, 0.326 mole of CO2 was used in the reaction.

Step 2:

The balanced equation for the reaction. This is given below:

6CO2 + 8H2O —> 2C3H7OH + 9O2

Step 3:

Determination of the number of mole of C3H7OH produced from the reaction. This is illustrated below:

From the balanced equation above,

6 moles of CO2 reacted to produce 2 moles of C3H7OH.

Therefore, 0.326 mole of CO2 will react to produce = (0.326 x 2)/6 = 0.109 mole of C3H7OH.

Step 4:

Conversion of 0.109 mole of C3H7OH to grams. This is illustrated below:

Number of mole of C3H7OH = 0.109 mole.

Molar mass of C3H7OH = (12x3)+ (7x1) + 16 + 1 = 60g/mol

Mass of C3H7OH =..?

Mass = mole x molar mass

Mass of C3H7OH = 0.109 x 60

Mass of C3H7OH = 6.54g.

Therefore, 6.54g of C3H7OH is produced from the reaction.

the answer is:  

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