Write your name here, show all calculations, and answer on the lines (6 questions).
QUIZ 4 – CHM 1211 – Summer 09
Name ________________________________________________________
1. What amount of bromine (in moles) reacts with 5.0 mol of aluminum to produce AlBr3? ___________
2 Al(s) + 3 Br2(g)
2 AlBr3(s)
2. What mass of sodium will react with 1.23 g of chlorine gas to produce sodium chloride? ___________
2 Na(s) + Cl2(g)
2 NaCl(s)
3. The reaction of excess NH3 and 18.3 g O2 produces 4.10 g NO. What is the percent yield of this reaction?
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
____________
4. For the following reaction, 18.71 grams of carbon monoxide are allowed to react with 27.18 grams of oxygen gas
according to the reaction below. What is the maximum amount of carbon dioxide that can be formed?
2 CO (g) + O2 (g)
2 CO2 (g)
______________
5. A solution is made by dissolving 19.0 g of methanol, CH3OH, in enough water to make exactly 500 mL of solution. What is the concentration (molarity) of CH3OH in mol/L?
____________
6. You need to make an aqueous solution of 0.120 M sodium hydroxide for an experiment in lab, using a 300 mL
volumetric flask. How much in grams of solid sodium hydroxide should you add?
____________
Write your name here, show all calculations, and answer on the lines (6 questions).
QUIZ 4 – CHM 1211 – Summer 09
Name _____KEY______________________________________________
1. What amount of bromine (in moles) reacts with 5.0 mol of aluminum to produce AlBr3? _7.5 mol Br2
2 Al(s) + 3 Br2(g)
2 AlBr3(s)
5.0 mol Al * 3 mol Br2 = 7.5 mol Br2
2 mol Al
2. What mass of sodium will react with 1.23 g of chlorine gas to produce sodium chloride? _0.800 g Na
2 Na(s) + Cl2(g)
2 NaCl(s)
1.23 g Cl2 * 1 mol Cl2 = 0.0174 mol Cl2
70.9 g Cl2
;
0.0174 mol Cl2 * 2 mol Na = 0.0348 mol Na
1 mol Cl2
0.0347 mol Na * 23 g Na = 0.800 g Na
1 mol Na
3. The reaction of excess NH3 and 18.3 g O2 produces 4.10 g NO. What is the percent yield of this reaction?
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
18.3 g O2 * 1 mol O2 = 0.572 mol O2
32 g O2
;
___29.8%___
0.572 mol O2 * 4 mol NO = 0.458 mol NO
5 mol O2
0.458 mol NO * 30 g NO = 13.74 g NO ; % yield = 4.10 g * 100 = 29.84%
1 mol NO
13.74 g
4. For the following reaction, 18.71 grams of carbon monoxide are allowed to react with 27.18 grams of oxygen gas
according to the reaction below. What is the maximum amount of carbon dioxide that can be formed?
2 CO (g) + O2 (g)
2 CO2 (g)
___29.39 g CO2___
18.71 g CO * 1 mol CO = 0.668 mol CO ; 27.18 g O2 * 1 mol O2 = 0.849 mol O2
28 g CO
32 g O2
Diving each amount by stoichiometric coefficient 0.668/2 = 0.334 and 0.849/1 = 0.849, the smallest number is 0.334
thus limiting reactant is CO, need to base calculations on 0.668 mol CO available. Also, to react 0.668 mol CO, 0.334
mol O2 is needed, and there is more available, so there is plenty of O2 to react all CO.
0.668 mol CO * 2 CO2 = 0.668 mol CO2
2 CO
; 0.668 mol CO2 * 44 g CO2 = 29.39 g CO2
1 mol CO2
5. A solution is made by dissolving 19.0 g of methanol, CH3OH, in enough water to make exactly 500 mL of solution.
What is the concentration (molarity) of CH3OH in mol/L?
1.19 M CH3OH
M = mol/ L ; 19.0 g CH3OH * 1 mol CH3OH = 0.594 mol CH3OH ; 500 mL * 1 L
32 g CH3OH
1000 mL
= 0.5 L
M = 0.584 mol CH3OH = 1.19 M CH3OH
0.5 L
6. You need to make an aqueous solution of 0.120 M sodium hydroxide for an experiment in lab, using a 300 mL
volumetric flask. How much in grams of solid sodium hydroxide should you add?
1.44 g NaOH
M = mol/L ; 300 ml * 1 L
1000 mL
= 0.3 L ;
0.12 M = X mol ;
0.3 L
0.036 mol NaOH * 40 g NaOH = 1.44 g NaOH
1 mol NaOH
X mol = 0.036 mol NaOH