If a plant has 88.0 g carbon dioxide and 64.0g water available for photosynthesis determine:
a the limiting reactant
b. the excess reactant and the mass in excess
c. the mass of glucose produced
a. The limiting reactant is the reactant that is completely consumed first, limiting the amount of product that can be formed. To determine the limiting reactant, we need to calculate the number of moles of each reactant and compare their stoichiometric ratios with the balanced equation for photosynthesis.
First, we need to convert the masses of carbon dioxide and water into moles. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and water (H2O) is 18.02 g/mol.
Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of CO2 = 88.0 g / 44.01 g/mol = 1.999 mol (approximately)
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 64.0 g / 18.02 g/mol = 3.553 mol (approximately)
Looking at the balanced equation for photosynthesis:
6 CO2 + 6 H2O → C6H12O6 + 6 O2
The stoichiometric ratio between CO2 and glucose is 6:1. Since we have 1.999 mol of CO2, we multiply it by the stoichiometric ratio:
1.999 mol CO2 * (1 mol glucose / 6 mol CO2) = 0.333 mol glucose (approximately)
The stoichiometric ratio between H2O and glucose is also 6:1. Therefore, with 3.553 mol of H2O, we multiply it by the stoichiometric ratio:
3.553 mol H2O * (1 mol glucose / 6 mol H2O) = 0.592 mol glucose (approximately)
Comparing the calculated moles of glucose, we see that the limiting reactant is CO2 since it produces a lower amount of glucose.
b. The excess reactant is the reactant that is not completely consumed and remains after the limiting reactant is completely reacted. To determine the mass of the excess reactant, we need to calculate the mass of the reactant that is in excess.
For the excess reactant, we need to determine how much of it is remaining based on the stoichiometry of the reaction when the limiting reactant is completely consumed. Since CO2 is the limiting reactant, it will be completely consumed, and the stoichiometric ratio between H2O and CO2 is 6:6. Therefore, for every 6 moles of CO2 consumed, 6 moles of H2O will be consumed.
Using the stoichiometric ratio, we can calculate the moles of H2O required to react with the available amount of CO2:
1.999 mol CO2 * (6 mol H2O / 6 mol CO2) = 1.999 mol H2O
Since we have 3.553 mol of H2O, the difference between the required moles and the available moles will give us the excess:
Excess H2O = 3.553 mol H2O - 1.999 mol H2O = 1.554 mol H2O
Finally, we can calculate the mass in excess:
Mass of excess H2O = excess H2O * molar mass of H2O
Mass of excess H2O = 1.554 mol * 18.02 g/mol = 27.98 g
Therefore, the excess reactant is water (H2O) with a mass of 27.98 g.
c. The balanced equation tells us that 6 moles of CO2 are needed to produce 1 mole of glucose. Since we have 1.999 mol of CO2 (the limiting reactant), we can calculate the moles of glucose produced:
1.999 mol CO2 * (1 mol glucose / 6 mol CO2) = 0.333 mol glucose (approximately)
To calculate the mass of glucose produced, we need to multiply the number of moles by the molar mass of glucose, which is 180.16 g/mol.
Mass of glucose produced = 0.333 mol * 180.16 g/mol = 59.98 g (approximately)
Therefore, the mass of glucose produced is approximately 59.98 g.
To determine the limiting reactant, we need to compare the number of moles of carbon dioxide and water.
First, let's find the number of moles for each reactant using their molar masses:
Molar mass of carbon dioxide (CO2) = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol
Number of moles of carbon dioxide = 88.0 g / 44.01 g/mol = 2.00 mol
Molar mass of water (H2O) = 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
Number of moles of water = 64.0 g / 18.02 g/mol = 3.55 mol
a. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. In this case, the reactant with fewer moles is the limiting reactant. Here, carbon dioxide (CO2) has 2.00 moles, while water (H2O) has 3.55 moles. Therefore, carbon dioxide is the limiting reactant.
b. To determine the excess reactant and the mass in excess, we need to calculate the amount of the excess reactant that is not consumed:
Number of moles of excess water = moles of reactant - moles of limiting reactant
Number of moles of excess water = 3.55 mol - 2.00 mol = 1.55 mol
Mass of excess water = number of moles of excess water * molar mass of water
Mass of excess water = 1.55 mol * 18.02 g/mol = 27.91 g
Therefore, water is the excess reactant with a mass of 27.91 grams.
c. To find the mass of glucose produced, we need to determine the stoichiometric ratio between the limiting reactant and the product (glucose). The balanced equation for photosynthesis is:
6 CO2 + 6 H2O -> C6H12O6 + 6 O2
From the equation, we can see that 6 moles of carbon dioxide are required to produce 1 mole of glucose.
Number of moles of glucose produced = moles of limiting reactant * (1 mole of glucose / 6 moles of carbon dioxide)
Number of moles of glucose produced = 2.00 mol * (1 mol / 6 mol) = 0.333 mol
To find the mass of glucose produced, we can use its molar mass:
Molar mass of glucose (C6H12O6) = 6 * 12.01 g/mol (C) + 12 * 1.01 g/mol (H) + 6 * 16.00 g/mol (O) = 180.18 g/mol
Mass of glucose produced = number of moles of glucose * molar mass of glucose
Mass of glucose produced = 0.333 mol * 180.18 g/mol = 60.08 g
Therefore, the mass of glucose produced is 60.08 grams.
To determine the limiting reactant, excess reactant, and the mass of glucose produced, we need to compare the available amounts of carbon dioxide and water and their respective stoichiometric ratios.
a) To find the limiting reactant, we need to calculate the amounts of each reactant in terms of moles. First, let's determine the number of moles of carbon dioxide and water.
The molar mass of carbon dioxide (CO2) is 44.01 g/mol, while the molar mass of water (H2O) is 18.02 g/mol.
Number of moles of carbon dioxide = mass of CO2 / molar mass of CO2
= 88.0 g / 44.01 g/mol
≈ 2.00 mol
Number of moles of water = mass of H2O / molar mass of H2O
= 64.0 g / 18.02 g/mol
≈ 3.55 mol
Now, let's examine the balanced chemical equation for the photosynthesis reaction:
6 CO2 + 6 H2O -> C6H12O6 + 6 O2
By comparing the coefficients in the balanced equation, we can see that the stoichiometric ratio of CO2 to H2O is 6:6 or 1:1. Therefore, we have an equal number of moles for both reactants (2.00 mol CO2 and 3.55 mol H2O), indicating that neither reactant is in excess.
b) Since there is no excess reactant, we do not need to determine the mass in excess.
c) To find the mass of glucose produced, we need to determine the limiting reactant. But since both reactants are the limiting reactants (in a 1:1 stoichiometric ratio), we simply consider one of the reactants.
From the balanced equation, we can see that for every 6 moles of CO2 (or H2O) used, 1 mole of glucose (C6H12O6) is produced. Therefore, the number of moles of glucose produced will be equal to the moles of water consumed (3.55 mol).
Mass of glucose produced = number of moles of glucose × molar mass of glucose
= 3.55 mol × 180.16 g/mol
≈ 638.38 g
Therefore, the mass of glucose produced is approximately 638.38 grams.
6CO2 + 6H2O ==> C6H12O6 + 6O2
mols CO2 = grams/molar mass
mols H2O = grams/molar mass
You need to redo the math because these are only approx but I obtained approx.
About 2 mols CO2; about 3.5
Convert 2 mols CO2 to mols glucose; that's 2 x (1 mol glucose/6 mol CO2) = about 0.33 mol glucose
Convert 3.5 mol H2O to mols glucose; that's about 3.5 x (1 mol glucose/5 mols H2O) = about 0.6.
You see the value is not the same for mols glucose which means one of the numbers is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. So CO2 is the limiting reagent. grams glucose = mols glucose x molar mass glucose = approx 0.33 x molar mass glucose = ?
For part b, the excess reagent obviously is H2O. How much. Use the same stoichiometry to determine how much H2O was used. That's approx 0.33 mols CO2 x (6 mols CO2/6 mol H2O) = about 0.33 mols H2O used.
You had approx 0.6 mol initially, you used about 0.33 so the difference is the amount H2O left in mols. To find grams, g = mols x molar mass.