# Stoichiometry

Discussion in 'DAT Discussions' started by Awuah29, Apr 14, 2007.

1. ### Awuah29 Christian predent

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How many moles of O2 are needed to react with 304.5 g of Cs2 in the combustion reaction below

Cs2(l) + 3o2 (g)---- Co2(g) + 2 So2 (g)

the ansewer is 12 mole, but I get 6 moles when I set it up.

How may grams of H2o will form when 10g of NH3 mixes with 10g of O2 according to the reaction below.

4 NH3(g) + 5 o2(g)----4 No2(g) + 6 H2o

3. ### skyisblue

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Hi Awuah, the answer key is right. It is 12 moles of O2 needed.

First, convert CS2 to moles and you get 4 moles of CS2. Then, you do a mole ratio of the reactants to get the number of moles of O2.

4 moles of CS2 x 3 moles of O2/1 mole of CS2 = 12 moles of O2.

It would help if you capitalize the elements accordingly.

4. ### Lonely Sol cowgoesmoo fan!

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Is the element, C and S (carbon and sulfur) or is it Cs (Caesium), if it is cesium then there is no way in hell it is 12, but if it C and S2 then it is 12. Yea, skyisblue is right, it must be C and S

5. ### orthdent786 Junior Member

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for your second question- i think it is a limiting reactant question:

1. 10 g NH3 X 1 mole NH3/ 17 g NH3 = 0.58 moles NH3
2. 10 g O2 X 1 mole O2/ 32 g O2 = 0.3125 moles O2

Since O2 has less moles it is the limiting reactant.

Therefore;

0.3125 moles O2 X (6 moles H20/ 7 moles O2) X (18 g H2O/ 1 mole H2O)=

4.82 g H2O

I think.. correct me if i'm wrong!

Balanced equation is;

4NH3 + 7O2 --> 4NO2 + 6H20

correct?

6. ### skyisblue

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the 2nd question is a limiting reactant problem, but the equation is not balanced.

Okay, with the balanced equation, I calculated it to be about 4.78 g of H2O.

4NH3 + 7O2 to 4NO2 + 6H2O

7. ### doc toothache

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If it was cesium there would be no SO2, and for that matter no CO2. The only product would be CsO2.

8. ### gochi

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Correct me if Im wrong, but arent we not allowed to use calculators ?

How am I suppose to do these difficult calculations w/o a calc ?

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