Can somebody help me with this?
How many grams of C2H5OH are used up when 6.25 grams of O2 react according to the following equation?
C2H5OH + 3O2 ---> 2CO2 + 3H2O
To solve this problem, we need to use the stoichiometry of the balanced chemical equation. Here's the step-by-step process:
Step 1: Find the molar mass of C2H5OH (ethanol).
The molar mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol. Adding these up, we have:
(2 * 12.01) + (6 * 1.01) + 16.00 = 46.08 g/mol
Step 2: Calculate the number of moles of C2H5OH.
We use the given mass of C2H5OH (6.25 g) and divide it by the molar mass we calculated in Step 1 to get moles:
6.25 g / 46.08 g/mol ≈ 0.1356 mol
Step 3: Use the stoichiometry of the balanced equation.
According to the balanced equation, 1 mole of C2H5OH reacts with 3 moles of O2. Therefore, 0.1356 moles of C2H5OH will react with:
0.1356 mol * (3 mol O2 / 1 mol C2H5OH) = 0.4068 mol O2
Step 4: Convert moles of O2 to grams.
Multiply the moles of O2 by its molar mass to get the mass in grams:
0.4068 mol * 32.00 g/mol = 13.01 g
Therefore, 13.01 grams of C2H5OH are used up when 6.25 grams of O2 react.
mols O2 = grams/molar mass = ?
mols O2 x 1/3 = mols C2H5OH used because the equation tells you 1 mol C2H5OH = 3 mols O2.
grams C2H5OH = mols x molar mass = ?