When methanol undergoes complete combustion, the products are carbon dioxide and water. 2CH3OH (l) + 3O2(g) --> 2CO2(g) + 4H2O (g). How much water ( in grams) can you maximally produce if you burn 50g of methanol? In real experiment, you isolate 10 mL of water (d=0.98 g/cm3) , what's your yield?
You must always have a balanced equation and you have that.
Convert 50 g methanol to moles. moles = grams/molar mass
Using the coefficients in the balanced equation, convert moles methanol to moles H2O.
Now convert moles H2O to grams. g = moles x molar mass. This is the theoretical yield.
The above is what you need to work any simple stoichiometric problem. The next step converts to yield.
%yield = (actual yield/theoretical yield)*100 = ??
You have actual yield in the problem of 10 mL water. Use the density to convert that to grams water, then substitute into the above percent formula.
To determine the maximum amount of water produced when burning 50g of methanol, we can use the stoichiometry of the balanced chemical equation.
The balanced equation is:
2CH3OH (l) + 3O2(g) --> 2CO2(g) + 4H2O (g)
First, let's calculate the molar mass of methanol (CH3OH):
C: 12.01 g/mol
H: 1.01 g/mol (there are 4 hydrogen atoms)
O: 16.00 g/mol
Molar mass of methanol:
(12.01 g/mol x 1) + (1.01 g/mol x 4) + (16.00 g/mol x 1) = 32.04 g/mol
Next, we need to determine the number of moles of methanol in 50g. We can use the formula:
moles = mass / molar mass
moles of methanol = 50g / 32.04 g/mol = 1.56 mol
According to the balanced equation, 2 moles of methanol produce 4 moles of water. Therefore, in this reaction, the stoichiometric ratio of methanol to water is 2:4.
Using this ratio, we can calculate the number of moles of water produced:
moles of water = (moles of methanol / 2) x 4 = (1.56 mol / 2) x 4 = 3.12 mol
Finally, we can convert the moles of water to grams using the molar mass of water (H2O), which is 18.02 g/mol:
mass of water = moles of water x molar mass of water = 3.12 mol x 18.02 g/mol = 56.36g
Therefore, the maximum amount of water that can be produced from burning 50g of methanol is 56.36 grams.
For the real experiment, if you isolate 10 mL of water with a density of 0.98 g/cm3, we can calculate the mass of water:
mass = volume x density = 10 mL x 0.98 g/cm3 = 9.8 grams
In this case, the yield of water is 9.8 grams, assuming all the water isolated is from the combustion of methanol.