Answer:
6.54g of C3H7OH
Step-by-step 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.