For the combustion of sucrose:
C12H22O11 + 12O2 → 12CO2 + 11H2O
There are 12.0 g of sucrose and 10.0 g of oxygen reacting in lab conditions.
How much H2O will be produced? (You will need notebook paper to solve this)
I don't have notebook paper but I suppose a computer will do.
This is a limiting reagent (LR) problem. You know that is the case when amounts are given for more than one reactant in an equation.
C12H22O11 + 12O2 → 12CO2 + 11H2O
mols sucrose = grams/molar mass = 12.0/342 = approx 0.035
How much H2O would this produce? That's 0.035 mols C12H22O11 x (11 mols H2O/1 mol C12H22O11) = 0.035 x 11 = approx 0.38 mols H2O.
mols O2 = 10.0/32 = 0.31
How much H2O would the O2 produce? That's 0.31 x (11 mols H2O/12 mol O2) = 0.28.
The smaller number is provided by O2; therefore, O2 is the LR.
grams H2O = mols H2O x molar mass H2O = 0.28 x 18 = ?
Check my arithmetic. Note that I've estimated all of the numbers. You should recalculate all of the steps and carry it out to 3 places. Post your work if you get stuck.
To determine the amount of H2O produced in the combustion of sucrose, we need to find the limiting reactant first. The limiting reactant is the one that gets completely consumed and determines the maximum amount of product formed.
Step 1: Calculate the molar masses of the reactants.
- Molar mass of C12H22O11 (sucrose) = 12(12.01) + 22(1.01) + 11(16.00) = 342.3 g/mol
- Molar mass of O2 = 2(16.00) = 32.00 g/mol
Step 2: Calculate the number of moles of each reactant.
- Moles of sucrose = mass/molar mass = 12.0 g/342.3 g/mol ≈ 0.0350 mol
- Moles of oxygen = mass/molar mass = 10.0 g/32.00 g/mol ≈ 0.3125 mol
Step 3: Determine the stoichiometry of the reaction.
From the balanced equation, we know that 1 mole of sucrose reacts with 12 moles of oxygen to produce 11 moles of water.
Step 4: Calculate the moles of water produced based on the limiting reactant.
To determine the limiting reactant, we compare the moles of each reactant with their stoichiometric ratio.
- Moles of water produced from 0.0350 mol of sucrose: 0.0350 mol × (11 mol H2O/1 mol sucrose) = 0.385 mol H2O
- Moles of water produced from 0.3125 mol of oxygen: 0.3125 mol × (11 mol H2O/12 mol O2) ≈ 0.2865 mol H2O
Since the moles of water produced from sucrose (0.385 mol) is greater than the moles of water produced from oxygen (0.2865 mol), the limiting reactant is oxygen.
Step 5: Calculate the mass of water produced.
The molar mass of H2O is 18.02 g/mol.
- Mass of water produced from 0.3125 mol of oxygen = 0.3125 mol × 18.02 g/mol = 5.63 g
Therefore, approximately 5.63 grams of H2O will be produced in the combustion of 12.0 g of sucrose and 10.0 g of oxygen.
To find out how much water (H2O) will be produced in the combustion of sucrose with the given amounts of sucrose and oxygen, you can follow these steps:
Step 1: Convert the given masses of sucrose (12.0 g) and oxygen (10.0 g) to moles.
To convert mass to moles, you'll need to use the molar mass of each substance. The molar mass of sucrose (C12H22O11) can be calculated by adding up the atomic masses of the elements present in its chemical formula.
The atomic masses:
- Atomic mass of carbon (C) = 12.01 g/mol
- Atomic mass of hydrogen (H) = 1.008 g/mol
- Atomic mass of oxygen (O) = 16.00 g/mol
Molar mass of sucrose (C12H22O11):
(12 carbon atoms x 12.01 g/mol) + (22 hydrogen atoms x 1.008 g/mol) + (11 oxygen atoms x 16.00 g/mol) = 342.34 g/mol
Now you can calculate the moles of sucrose and oxygen:
Moles of sucrose = Given mass of sucrose / Molar mass of sucrose
Moles of oxygen = Given mass of oxygen / Molar mass of oxygen
Step 2: Determine the mole ratio between sucrose and water in the balanced equation.
From the balanced equation:
C12H22O11 + 12O2 → 12CO2 + 11H2O
You can observe that for every mole of C12H22O11 (sucrose), 11 moles of H2O (water) are produced.
Step 3: Calculate the moles of water produced.
Multiply the moles of sucrose by the mole ratio of water to sucrose from the balanced equation.
Moles of water = Moles of sucrose x (Mole ratio of water to sucrose)
Step 4: Convert moles of water to grams.
Moles to grams conversion can be done by multiplying the moles of water by the molar mass of water (H2O).
Molar mass of water (H2O) = (2 x 1.008 g/mol) + (16.00 g/mol) = 18.02 g/mol
Grams of water = Moles of water x Molar mass of water
Now you can calculate the answer by following these steps on notebook paper with the given values of 12.0 g of sucrose and 10.0 g of oxygen.