1. KEY
GENERAL CHEMISTRY-I (1411)
S.I. # 7
1. Define the following:
a. Combination Reaction: two reactants combine to form one product
b. Decomposition Reaction: a single reactant forms 2 or more products
c. Combustion Reaction: when oxygen and a hydrocarbon react to form
water and carbon dioxide
2. Define the following:
a. Limiting Reactant: is completely consumed in a reaction.
b. Theoretical Yield: is the quantity of product calculated to form when all of
the limiting reagent reacts.
c. Percent Yield: actual yield divided by the theoretical yield multiplied by
100%.
3. The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and
CO2:
C6H12O6 (aq)  2 C2H5OH (aq) + 2 CO2 (g)
a. How many moles of CO2 are produced when 0.400 mol of C6H12O6 reacts in
this fashion?
0.400 mol C6H12O6 (2 mol CO2 / 1 mol C6H12O6) = 0.800 mol CO2
b. How many grams of C6H12O6 are need to form 7.50 g of C2H5OH?
7.50 g C2H5OH x (1 mol C2H5OH) x (1 mol C6H12O6) x (180.2 g C6H12O6) =
(46.07 g C2H5OH) (2 mol C2H5OH) (1 mol C6H12O6)
14.7 g C6H12O6
c. How many grams of CO2 form when 7.50 g of C2H5OH are produced?
7.50 g C2H5OH x (1 mol C2H5OH) x (2 mol CO2) x (44.01 g CO2) =
(46.07 g C2H5OH) (2 mol C2H5OH) (1 mol CO2)
7.16 g CO2
4. Why are the amounts of products formed in a reaction determined only by the
amount of the limiting reactant?
Because it is completely used up during the reaction; no more product can be
made when one of the reactants is unavailable.

2. KEY
5. Sodium hydroxide reacts with carbon dioxide as follows:
2 NaOH (s) + CO2  Na2CO3 (s) + H2O (l)
Which reagent is the limiting reactant when 1.85 mol NaOH and 1.00 mol CO2 are
allowed to react? How many moles of the excess reactant remain after the
completion of the reaction?
1.85 mol NaOH x (1 mol Na2CO3) = 0.925 mol Na2CO3 can be produced
(2 mol NaOH)
Since there is a 1:1 ratio for Na2CO3 : CO2 we can say 0.925 mol Na2CO3 produced
requires 0.925 mol CO2 to be consumed. Or 1 mol of CO2 initially – 0.925 mol CO2
reacted = 0.075 mol CO2 remain.
Check by doing the following table
2 NaOH (s) + CO2  Na2CO3 (s) + H2O (l)
Initial 1.85 mol 1.00 mol 0 mol
Change -1.85 mol -0.925 mol +0.925 mol
Final 0 mol 0.075 mol 0.925 mol
6. When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is
obtained:
C6H6 + Br2  C6H5Br + HBr
a. what is the theoretical yield of bromobenzene in this reaction when 30.0 g of
benzene reacts with 65.0 g of bromine?
a. Determine the LR and the max amount of product it could produce then the % yield.
30.0 g C6H6 x (1 mol C6H6) x = 0.384 mol C6H6
(78.11 g C6H6)
65.0gBr2 x (1 mol Br2) = 0.407 mol Br2
(159.8g Br2)
since C6H6 and Br2 react in a 1:1 mol ratio, C6H6 is the limiting reactant and
determines the theoretical yield.
0.3841 mol C6H6 x (1mol C6H5 Br) x (157.0g C6H5Br) = 60.3 g C6H5Br = T.Y.
(1 mol C6H6) (1mol C6H5 Br)
b. If the actual yield of bromobenzene was 56.7 g, what was the percentage yield?
% yield = (56.7 g C6H5Br actual) x 100% = 94.0%
(60.3 g C6H5Br theoretical)