Answer :
Answer:
0.17 mole of glucose is formed.
Explanation:
Step 1:
The equation for the reaction. This is given below:
CO2 + H2O —> C6H12O6 + O2
Step 2:
Balancing the equation.
The equation can be balanced as follow:
CO2 + H2O —> C6H12O6 + O2
There are 6 atoms of C on the right side and 1 atom on the left side. It can be balance by putting 6 in front of CO2 as shown below:
6CO2 + H2O —> C6H12O6 + O2
Therefore are 12 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 6 in front of H2O as shown below:
6CO2 + 6H2O —> C6H12O6 + O2
There are a total of 8 atoms of O the right side and a total of 18 atoms on the left. It can be balance by putting 6 in front of O2 as shown below:
6CO2 + 6H2O —> C6H12O6 + 6O2
Now the equation is balanced.
Step 3:
Determination of the number of mole of glucose (C6H12O6) produced by 1 mole of water.
This is illustrated below:
6CO2 + 6H2O —> C6H12O6 + 6O2
From the balanced equation above,
6 moles of H2O produced 1 mole of C6H12O6.
Therefore, 1 mole of H2O will produce = 1/6 = 0.17 mole of C6H12O6.
Answer:
IF 1.0 mol of water is available, 0.167 moles of glucose (C6H12O6) will be produced
Explanation:
Step 1: Data given
Number of moles of water = 1.0 moles
Carbon dioxide is in excess
Step 2: The balanced equation
6CO2 + 6H2O → C6H12O6 + 6O2
Step 3: Calculate moles of glucose
For 6 moles CO2 we need 6 moles H2O to produce 1 mol glucose and 6 moles O2
For 1.0 moles of water we'll have 1.00 / 6 = 0.167 moles of glucose
IF 1.0 mol of water is available, 0.167 moles of glucose (C6H12O6) will be produced