more mole ratios =/?

Question

im so confsed on this... i would like no know how to set up proportions to solve her...

1.) C12H22011 (sucrose) + 302 ---> 2H3C6H5O7 + 3H20
H3C6H5O7 = Citric Acid
Determine the mass of citric acid produced when 2.50 mol C12H22011 is produced.

2.) Determine the amount of ethyl butanote produced ig 4.50 mol ethanol is used.
C2H5OH + C3H7COOH -----> NC3H7COOC2H5 + H20
(ethyl butanote = C3H7COOC2H5) (alcohol ethanol= C2 H5OH)
(butanic acid C3H7COOH)
Determine the mass if ethyl butanoate produced if 4.50 mol of ethanol is used.

***** i would really appreciate if you could explain to me the SET UP of these problems so i can try to figure them out. thanks so much.

Answer 1

1) 1 mole of sucrose produces 2 moles of citric acid.

So if you say the you are getting x moles of citric acid you will have to start with (1/2)x mole of sucrose.

mass of sucrose = 1x/2 moles sucrose * (gram sucrose/ mole sucrose)

2) here it's easier it's a 1:1 relationship

1 mole ethanol will give 1 mole of ethylbutanoate

x moles ethanol gives x moles ethylbutanoate

mass EtButanoate= x * (g Etbutanoate/mole Etbutanoate)

Answer 2

Before I answer your questions, here are some useful tips:

- When balancing chemical equations and calculating mole ratios for reactants and products, it usually helps to group all 'like' atoms together when writing out formulas, which is how I've written them below.

- Once an equation is balanced, the numbers in front of each substance represent the MOLES (not mass!!!) that are either consumed (if it's a reactant/on the left side) or produced (if it's a product/on the right side) by the reaction....when there is no coefficent written, it's understood to be "1"

- Molecular weights for the compounds in question will be needed whenever you have to convert from moles into mass
(MW's can be looked up online or in reference books easily enough, but even if you have to figure them out by hand, a periodic table is all you need)

Okay, here we go.............

1. C12H22O11 (sucrose) + 3 O2 --> 2 C6H8O7 (CA) + 3 H2O

In this equation, 2 moles of citric acid are produced for every mole of sucrose that's consumed. This means that if 2.5 moles of sucrose are used up, then 5 (2*2.5) MOLES of citric acid will be created. Since the problem asks for the MASS instead, the only other information you'll need is the MW for citric acid, which is 192.13 g/mol*. 5 moles of CA =(5 x 192.13) g = 960.65 g


2. C2H6O (ethanol) + C4H8O2 -----> C6H12O2 (EB) + H20

In this equation, there are no coefficients written, so all of the molar ratios are 1:1. The problem specifies that 4.5 moles of ethanol are consumed, which means that 4.5 moles of ethyl butanoate (also known as ethyl butyrate) are created. Since the MW for this substance is 116.16 g/mol**, the mass of EB produced = (4.5 x 116.16) g = 522.72 g

I hope this helps..... :-)

Answer 3

It's hard for me to type these out the way they should be... but hopefully this will be useful.

Number 1)

2.50 mol C12H22O11 x (2 mol H3C6H5O7/ 1 mol C12H22O11) x (192.1 g H3C6H5O7/ 1 mol H3C6H5O7)
= 960.5 g H3C6H5O7

Explanation (that I hope makes sense):
Always start out with what you know, and think about where you need to get to.
Here, we have 2.50 moles of C12H22011 (sucrose). We need to get to grams of H3C6H5O7 (citric acid).

From the balance equation, we know that 1 mole of sucrose produces 2 moles of citric acid. Multiplying 2.50 moles of sucrose by (2 mol H3C6H5O7/ 1 mol C12H22O11) This cancels sucrose out of the equation, so now you know how many moles of citric acid you have.

Now we can find how many grams of citric acid is produced by using the molar mass of citric acid, which is 192.1 grams per mole. Once again, multiply so that moles cancel out, and you're left with grams.

Problem 2)
C2H5OH + C3H7COOH -----> NC3H7COOC2H5 + H20
Is that N supposed to be there? It doesn't make sense for it to be...
I'm going to assume that's a typo... in which case the real equation would be:
C2H5OH + C3H7COOH -----> C3H7COOC2H5 + H20
This equation is already balanced, and you work it out like you did number one.

4.50 mol C2H5OH x (1 mol C3H7COOC2H5 / 1mol C2H5OH) x
(116.2 g C3H7COOC2H5 / 1 mol C3H7COOC2H5)
= 522.9 grams of C3H7COOC2H5