Methanol burns in oxygen to produce carbon (iv) oxide and water if 9 grams of methanol are used in the combustion process what mass of water is produced
write the balanced equation.
determine how many moles of methanol you have in nine grams.
Now, look at the coefficents in the balanced duquation. what is the mole methanol: mole water ratio?
convert the moles of water then to grams.
To calculate the mass of water produced during the combustion of methanol, we need to use the balanced chemical equation for the reaction.
The balanced equation for the combustion of methanol is:
2CH3OH + 3O2 -> 2CO2 + 4H2O
From the balanced equation, we can see that for every 2 moles of methanol (CH3OH) burned, 4 moles of water (H2O) are produced.
Step 1: Convert mass of methanol (CH3OH) to moles.
To do this, we need to know the molar mass of methanol. The molar mass of methanol (CH3OH) is calculated by adding the atomic masses of its constituent atoms: 1 carbon (C), 4 hydrogens (H), and 1 oxygen (O).
Molar mass of CH3OH = (1*12.01 g/mol) + (4*1.01 g/mol) + (1*16.00 g/mol) = 32.04 g/mol
Using the molar mass, we can calculate the number of moles of methanol:
Number of moles of methanol = mass of methanol / molar mass of CH3OH
Number of moles of methanol = 9 g / 32.04 g/mol
Step 2: Calculate the moles of water produced.
From the balanced equation, we know that for every 2 moles of CH3OH burned, 4 moles of H2O are produced. Therefore, the moles of water produced will be twice the moles of methanol used.
Moles of water produced = (Number of moles of methanol) * 2
Step 3: Convert moles of water to mass.
To calculate the mass of water produced, we multiply the moles of water by the molar mass of water (H2O).
Mass of water produced = (Moles of water produced) * (molar mass of H2O)
Molar mass of H2O = (2 * 1.01 g/mol) + (1 * 16.00 g/mol) = 18.02 g/mol
Mass of water produced = (Moles of water produced) * 18.02 g/mol
Now, we can plug in the values:
Number of moles of methanol = 9 g / 32.04 g/mol = 0.28 mol
Moles of water produced = 0.28 mol * 2 = 0.56 mol
Mass of water produced = 0.56 mol * 18.02 g/mol = 10.11 grams
Therefore, 10.11 grams of water will be produced when 9 grams of methanol is burned.