Chemistry 101 Sections 501-512 Worksheet With Answer Key - Texas A And M University, 2001 Page 2
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1
2NAME: _________________________________
Form 2 (green)
September 17, 2001
QUIZ #1
CHEM 101
1) Fill in the blanks to balance the following equations:
+ ____C ! __2_FeCl
__2_FeTiO
+ _2__Cl
+ __2_TiO
+ ____CO
3
2
2
2
2
2) Name the following compounds:
a) MgBr
__magesium bromide_______
2
a) FeCl
___iron(III) chloride_____
3
c) N
H
____dinitrogen tetrahydride______
2
4
3) Salicylic acid, or aspirin, is 60.87% C, 4.38% H, and 34.65% O by mass.
a) What is the empirical formula for this compound? Assume 100 g
! ! ! ! 5.068/2.166 = 2.340 = 7/3
C: (60.87 g C) x (1 mol C/12.011 g C) = 5.068 mol C
! ! ! ! 4.35/2.166 = 2.01 = 2
H: (4.38 g H) x (1 mol H/1.008 g H) = 4.35 mol H
! ! ! ! 2.474/2.166 = 1.000 = 1
O: (34.65 g N) x (1 mol O/15.999 g O) = 2.166 mol O
O ! ! ! ! need whole numbers so mulitply by 3 ! ! ! ! C
C
H
H
O
7/3
2
7
6
3
b) If the molar mass of the compound is 138.12 g/mol what is the molecular formula of the
compound?
Using empirical formula molar mass of C
H
O
= (7 x 12.011 g/mol) + (6 x 1.008 g/mol)
7
6
3
+ (3 x 15.999 g/mol) = 138.12 g/mol
Molar mass/Emprical molar mass = 138.12/138.12 = 1.000 = 1
Molecular formula is same as empirical formula, C
H
O
7
6
3
3) Diborane, B
H
, is a valuable compound in the synthesis of new organic compounds. One
2
6
way this compound can be made is by the reaction shown below.
! B
2NaBH
+ I
H
+ 2NaI + H
4
2
2
6
2
A reaction takes place between 1.23 g of NaBH
and 4.57 g of I
and goes to completion.
4
2
a) What is the limiting reagent (circle one shown to the right):
NaBH
I
4
2
SHOW WORK!!
NaBH
: 1.23 g NaBH
x {1 mol/[22.990 g + 10.811 g + 4(1.008 g)]} = 0.0325 mol NaBH
4
4
4
I
: 4.57 g I
x [1 mol/(2 x 126.904)]= 0.0180 mol I
2
2
2
! ! ! ! Have that much
Need 0.0325 NaBH
x (1 mol I
/2 mol NaBH
) = 0.0163 mol I
4
2
4
2
! ! ! ! Do not have that much
Need 0.0180 mol I
x (2 mol NaBH
/2 I
) = 0.0360 mol NaBH
2
4
2
4
Thus, NaBH
is limiting reagent
4
b) Calculate the maximum yield of B
H
.
2
6
0.0325 mol NaBH
x (1 mol B
H
/2 mol NaBH
) = 0.0163 mol B
H
4
2
6
4
2
6
0.0163 mol B
H
x {[(2 x 10.811 g) + (6 x 1.008 g)]/1 mol B
H
} = 0.451 g B
H
2
6
2
6
2
6
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