Answer to group practice chap3
1.
2. What is the mass percent (%) for O in SO2?
(a) 38.09
(b) 45.41
(c) 50.00
(d) 53.86
(e) 56.43
Hint: p. 87. Equation (3.1). Example 3.8. % S = (mass of 2 O/ mass of SO2) x 100% =
{2x16/(1x32+2x16)}x100% = 50.0%
3.
4. Chemical analysis shows the composition of a compound containing carbon and hydrogen, to be 80.00% carbon
and 20% hydrogen and the molar mass is 30 g. What is its molecular formula?
(a) CH
(b) C2H4
(c) C2H6
(d) C6H12 (e) C10H22
Hint: p.p. 91-92. Example 3.11 or p. 109 (3.54). The molecular formula is an integral multiple of empirical
formula. That is, the molar mass = empirical molar mass x integer. From C: H = 80.00/12 : 20.00/1 = 6.66: 20 =
1: 3. So the empirical formula is CH3 and the empirical molar mass of CH3 = 12x1+1x3 =15. So the integer =
30/15 = 2. Thus there are two empirical formulas in a molecular formula. Therefore, the molecular formula is
C2H6.
5.
6. In the reaction of Al(OH)3 with H2SO4, how many moles of water can be produced If the reaction is begun
with 5.500 mole of Al(OH)3?
2Al(OH)3 + 3H2SO4  Al2(SO4)3 + 6H2O
(a) 2.50
(b) 4.75
(c) 6.32
(d) 7.58
(e) 16.50
Hint: p.p. 97-101. Examples 3.13. and 3.14. Figure 3.8. Stoichiometrey. Method One (ratio approach): From
the equation, the involving species, 2 Al(OH)3 and 6 H2O with coefficients 2 and 6 respectively, tell us that for 2
moles of Al(OH)3 it produces 6 moles of H2O. According to this proportion or ratio, 5.500 mole of Al(OH) 3
requires 5.500 x (6/2) = 16.50 moles of H2O.
Method Two (road map approach): Apply the road map or say the dimensional analysis. The road map is
grams of substance A  moles of substance A moles of substance B  grams of substance B.
Note:
(1) substance A is the one with given (or known) information of mass (or mole); substance B is
the one needed to be calculated.
(2) grams of substance A  molar mass of substance A = moles of substance A.
(3) moles of substance A  coefficient of substance B  coefficient of substance A = moles of
substance B.
(4) grams of substance B = moles of substance B  molar mass of substance B.
Since this question starts at moles and thus we only apply (2):
5.500 moles Al(OH)3 x {6 H2O/2 Al(OH)3} = 16.50 moles H2O
7. How many grams of H2O could be formed by the reaction of 16.0 g of CH4 with 48.0 g of O2?
CH4 + 2O2  CO2 + 2H2O
(a) 27.0
(b) 37.3
(c) 46.8
(d) 54.1
(e) 58.7
Hint: p.p. 97-101. Examples 3.13. and 3.14. Figure 3.8. Stoichiometrey. This question provides two known
quantities of substance A and thus this is the limiting reagent question. So we need to apply two times of allfour-step-road map to figure out what is the true mass of water. Note that the limiting reagent limits the
(maximum) of product(s) that can be produced. In the road map application, the one produces the smallest
amount of the product is the true limiting reagent. The one with the left over is the excess reagent.
Assume CH4 (i. e. substance A # 1) is the limiting reagent:
÷ 16 g/mol
x 2H2O/1CH4
x 18 g/mol
16 g CH4 ------------- 1 mole CH4 ---------------- 2 mole H2O ------------- 36 g H2O
Assume O2 is the limiting reagent:
÷ 32 g/mol
x 2H2O/2O2
x 18 g/mol
48 g O2 ------------- 1.5 mole O2 ---------------- 1.5 mole H2O------------- 27 g H2O
Because O2 produces the least amount of H2O (can compare by using mole, 1.5 < 2, or grams, 27 < 36), it is the
true limiting reagent and the maximum amount of water produced is 27 grams. The CH 4 is the excess reagent.
8. What is the excess reagent in the above reaction (Q.23)of CH4 and O2? How many grams of the excess
reagent were consumed?
(a) CH4, 12 grams
(b) O2, 12 grams
(d) O2, 4 grams
(e) CH4, 8 grams
(c) CH4, 4 grams
Hint: p.p. 101-103. Example 3.15. Now the substance A is O 2 and CH4 is the substance B because we need to
calculate its amount consumed stoichiometrically. Apply the road map:
÷ 32 g/mol
x 1CH4/2O2
x 16 g/mol
48 g O2 ------------- 1.5 mole O2 ---------------- 0.75 mole CH4------------- 12 g CH4
Thus, there are (16 – 12) = 4 grams of CH4 left.
This is because # grams excess = # grams original – # grams used.
Or say # moles excess = # moles original – # moles used.
9. % yield = (actual yield/ theoretical yield) x 100%
Now from the road map, we calculate the theoretical yield (note that since the actual yield
given is in unit of grams, so we need to complete the entire road map).
÷ 78 g/mol
x 1C6H5NO2 /1C6H6
x 123g/mol
31.2 gC6H6------------ 0.4mole C6H6 ----------------------- 0.4mole C6H5NO2 --------- 49.2 g C6H5NO2
Thus the percent yield = {38.7 g / 49.2 g} x 100% = 78.7%