C6H12O6(s) +6CO2(g) +6H2O9g)
1) given 2 moles of glucose and 14 moles of O2, how many moles of CO2 could be produced?
2) which is the limiting reactant ?
3) how many moles of the excess reactant remain?
To answer these questions, we need to start by balancing the chemical equation for the combustion of glucose:
C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(g)
Now we can proceed to answer each question:
1) Given 2 moles of glucose and 14 moles of O2, we can use the balanced equation to determine the stoichiometric ratio between glucose and CO2. From the equation, we see that 1 mole of glucose produces 6 moles of CO2. Therefore, by using a simple ratio calculation, we can find the number of moles of CO2 produced:
2 moles of glucose * (6 moles of CO2 / 1 mole of glucose) = 12 moles of CO2
So, 12 moles of CO2 could be produced.
2) To determine the limiting reactant, we compare the number of moles of glucose and oxygen (O2) used in the reaction. From the given information, we have 2 moles of glucose and 14 moles of O2. The balanced equation tells us that 1 mole of glucose reacts with 6 moles of O2. So, if we convert the moles of glucose to moles of O2 using the stoichiometric ratio, we have:
2 moles of glucose * (6 moles of O2 / 1 mole of glucose) = 12 moles of O2
Comparing this to the 14 moles of O2 provided, we can see that O2 is the limiting reactant because only 12 moles of O2 are required compared to the excess 14 moles that are available.
3) To find the moles of the excess reactant remaining, we need to calculate how much of the non-limiting reactant is left after the reaction. In this case, glucose is the non-limiting reactant. We know that 2 moles of glucose were initially used, and since we determined that O2 is the limiting reactant, we can calculate how many moles of glucose are consumed using the stoichiometric ratio:
12 moles of O2 * (1 mole of glucose / 6 moles of O2) = 2 moles of glucose consumed
Subtracting this from the initial 2 moles of glucose, we find that no moles of glucose remain. Thus, the moles of the excess reactant (glucose) remaining is 0.