If you reacted 12.0 g of C7H8O2 (124.14 g/mol) with 200.0 g of O2, (31.998 g/mol) what mass (in g) of H2O (18.015 g/mol) can you theoretically make?
convert to moles
write the equation -- it will tell you how many moles of H2O to expect
convert back to grams.
To determine the mass of H2O that can be theoretically made from the reaction between 12.0 g of C7H8O2 and 200.0 g of O2, we first need to identify the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed in the reaction, determining the maximum amount of product that can be formed.
1. Calculate the number of moles for each reactant:
Number of moles of C7H8O2:
12.0 g / 124.14 g/mol = 0.09665 mol
Number of moles of O2:
200.0 g / 31.998 g/mol = 6.250 mol
2. Use the stoichiometry of the balanced chemical equation to determine the mole ratio between C7H8O2 and H2O:
The balanced equation for the reaction is:
C7H8O2 + 9O2 -> 7CO2 + 4H2O
From this equation, we can see that each mole of C7H8O2 produces 4 moles of H2O.
3. Calculate the maximum number of moles of H2O that can be formed:
Maximum number of moles of H2O = 0.09665 mol x (4 mol H2O / 1 mol C7H8O2) = 0.3866 mol
4. Calculate the mass of H2O using the molar mass:
Mass of H2O = 0.3866 mol x 18.015 g/mol = 6.965 g
Therefore, the theoretical mass of H2O that can be produced in this reaction is 6.965 g.
To find the mass of H2O that can be theoretically made, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thus limiting the amount of product that can be formed.
To determine the limiting reactant, we can use the concept of stoichiometry and compare the amounts of reactants to their respective ratios in the balanced chemical equation.
The balanced chemical equation for the reaction is:
C7H8O2 + 9O2 -> 7CO2 + 4H2O
First, let's calculate the number of moles for each reactant:
Moles of C7H8O2 = mass of C7H8O2 / molar mass of C7H8O2
Moles of C7H8O2 = 12.0 g / 124.14 g/mol = 0.0968 mol
Moles of O2 = mass of O2 / molar mass of O2
Moles of O2 = 200.0 g / 31.998 g/mol = 6.250 mol
Next, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation. According to the stoichiometric ratio, 1 mole of C7H8O2 reacts with 9 moles of O2 to produce 4 moles of H2O.
Using the mole ratio between C7H8O2 and O2:
Moles of O2 required = moles of C7H8O2 x (9 moles O2 / 1 mole C7H8O2)
Moles of O2 required = 0.0968 mol x 9 = 0.8712 mol
Since we have 6.250 moles of O2 available and we need only 0.8712 moles, it means that O2 is in excess and C7H8O2 is the limiting reactant.
Now, we can calculate the moles of H2O that can be theoretically formed from the limiting reactant:
Moles of H2O = moles of C7H8O2 x (4 moles H2O / 1 mole C7H8O2)
Moles of H2O = 0.0968 mol x 4 = 0.3872 mol
Finally, let's convert the moles of H2O to grams:
Mass of H2O = moles of H2O x molar mass of H2O
Mass of H2O = 0.3872 mol x 18.015 g/mol = 6.975 g
Therefore, the theoretically maximum mass of H2O that can be produced in this reaction is approximately 6.975 grams.