how do i do this chemistry problem?

Question

the combustion of methanol produces carbon dioxide gas and water vapor.
2CH3OH(l) + 3O2(g) --> 2CO2(g) + 4H2O(g)
what mass of water vapor forms when 52.4g of methanol burn?

there are a lot of these types of questions on my homework and i don't understand how to do them. can someone please show me how step by step? thanks

Answer 1

u can ask ur teacher!

Answer 2

1. find the number of moles of methanol. this is done by dividing Mass by its Mr. The Mr is the relative formula mass.
For methanol, Mr= 12+3(1)+16+1=32 (another answer said that this was 36, which is wrong, as you can see in wikipedia if you type in methane, otherwise his working is correct)
therefore: 52.4/32 = 1.6375 moles of Methanol

then you look at the ratio of methanol to water which is 2:4 or 1:2 so you times that value by 2.

1.6375 x 2 = 3.275 moles of water

times that value by the Mr of water (reversing what was done previously but for water) will give you the mass.
therefore, 3.275 x 18 = 58.95g of water.

Answer 3

The balanced equation is the first clue. The equation shows that 2 molecules of CH3OH (methanol) will produce 4 molecules of H2O (water). A molecule is a very small amount of a compound. It is much more convenient to consider moles (1 mole = 1 gram molecular weight).

The problem tells you that you started with 52.4 g of methanol. How many moles did you start with? 1 mole of CH3OH weighs 32 g, so you started with 52.4 / 32 moles of methanol (1.6375 moles of methanol).

If (from the balanced equation) 2 moles of CH3OH produces 4 moles of H2O, how many moles of H2O will 1.6375 moles of CH3OH produce? 4 x (1.6375 / 2.00) or 3.275 moles of H2O.

1 mole of water weighs 18 grams, so you will get 3.275 x 18 grams of water (3.275 x 18 = 58.95).

Start with a balanced equation.
Only 2 parts of the equation are needed in this problem-
2CH3OH = 4H2O
The grams of methanol divided by 2 times the MW of methanol equals the grams of water (unknown) divided by 4 times the MW of water (where 2 and 4 are the coefficients of the balanced equation).

Answer 4

Convert the mass in grams of methanol to moles of methanol. Use a periodic chart to find the MOLECULAR MASS (g/mol) of methanol. Once you find the moles of methanol, you see that there are 2 moles of water vapor produced for every mol of methanol combusted (4/2 ratio). Then you find the number mol of water vapor formed. Multiply this by the molecular weight of water vapor. I'l do this example

C: 12g/mol
H: 1G/mol
O: 16g/mol

Methanol: 34g/mol

52.4g*1mol/34g*2molH2O(g)/mol Methanol*18gH2O/mol=55.5grams!

Answer 5

You need to follow the true path to chemical correctness: Mass to moles to moles to mass.

An example. Find the mole wt of methanol[32].
Divide the mass by the mole wt. This gives you the MOLES of methanol in the reaction.
Now check the reaction. For each TWO MOLES of methanol, you form FOUR MOLES of water, or a 1:2 ratio. So double the moles of methanol and you have the moles of water.
Now find the mole wt of water [18]
Multiply that by the moles of water.
You have travelled the true path.

Answer 6

2 CH3OH + O2 ---> 2CO2 +4 H2O we calculate the molecular mass of each compound C
mCH3OH =12+16+4=32
mH2O = 2+16=18

so if you look the component of interest you see that

2*32g = 64g of CH3OH correspond to 4*18 =72 g of H2O

and so 52.4 g of CH3OH gives 52.4*72/64 = 58.95g of H2O