176
Stotehlometry II Calculations Based on Chemical Equations
(a) Weight of pure NaCl = (0 0392 moles NaCl) sS 5
= 2 29 g NaCl
(b) If x = the required grams of 90 0% pure NaMnO 4 , you can see that 0 90* g
must contain 0 00784 moles of NaMnO 4 That is,
0 90* g = ( 00784 moles NaMnO 4 )( 141 9
= 1 11 g pure NaMnO 4
/
gNaMnQ 4 \
I,
mole NaMnOj
x = —— = 1 23 g impure NaMnO 4
PROBLEM:
(a) What volume of oxygen at 20°C and 750 torr is needed to burn 3 00 liters of
propane, C 3 H 8 , also at 20°C and 750 torr
9
(b) What volume of air (21 0% O 2 by volume) would be required under the same
conditions
9 The products of combustion are solely CO 2 and H 2 O
SOLUTION:
First, you must have a balanced chemical equation on whic'h to base your calculation Because the three C atoms of C 3 H 8 are converted to 3CO 2 , and the 8H atoms
are converted to 4H 2 O, you can readily see that the 10 oxygen atoms needed in
this much CO 2 and H 2 O must come from 50 2 Therefore,
C 3 H 8 + 50 2 4. 3CO 2 + 4H 2 O
Second, you must realize that, when the two substances you are interested in are
both gases you can make a much simpler calculation than that involved in the
"three simple steps " You recall (see p 160) that equal volumes of gases under the
same conditions of temperature and pressure contain the same number of moles
(or molecules) The chemical equation shows that you need 5 moles of O 2 per mole
of C 3 H 8 , therefore, you will need 5 times the volume of O 2 as the volume of C 3 H 8
under the same conditions Therefore,
(a) Volume of O 2 = ( ,
5 m °
le *°
2 ) (3 00 liters C 3 H 8 )
\ 1 mole L. 3 rig/
= 150 liters of O 2
(b) If V = the required volume of air (also a gas) that is 21 0% O 2 , you can see that
0 210V liters of air must provide 15 0 liters of O 2 That is,
0 210V = 15 0 liters O 2
V = ;prr7r =714 liters of air
Stotehlometry II Calculations Based on Chemical Equations
(a) Weight of pure NaCl = (0 0392 moles NaCl) sS 5
= 2 29 g NaCl
(b) If x = the required grams of 90 0% pure NaMnO 4 , you can see that 0 90* g
must contain 0 00784 moles of NaMnO 4 That is,
0 90* g = ( 00784 moles NaMnO 4 )( 141 9
= 1 11 g pure NaMnO 4
/
gNaMnQ 4 \
I,
mole NaMnOj
x = —— = 1 23 g impure NaMnO 4
PROBLEM:
(a) What volume of oxygen at 20°C and 750 torr is needed to burn 3 00 liters of
propane, C 3 H 8 , also at 20°C and 750 torr
9
(b) What volume of air (21 0% O 2 by volume) would be required under the same
conditions
9 The products of combustion are solely CO 2 and H 2 O
SOLUTION:
First, you must have a balanced chemical equation on whic'h to base your calculation Because the three C atoms of C 3 H 8 are converted to 3CO 2 , and the 8H atoms
are converted to 4H 2 O, you can readily see that the 10 oxygen atoms needed in
this much CO 2 and H 2 O must come from 50 2 Therefore,
C 3 H 8 + 50 2 4. 3CO 2 + 4H 2 O
Second, you must realize that, when the two substances you are interested in are
both gases you can make a much simpler calculation than that involved in the
"three simple steps " You recall (see p 160) that equal volumes of gases under the
same conditions of temperature and pressure contain the same number of moles
(or molecules) The chemical equation shows that you need 5 moles of O 2 per mole
of C 3 H 8 , therefore, you will need 5 times the volume of O 2 as the volume of C 3 H 8
under the same conditions Therefore,
(a) Volume of O 2 = ( ,
5 m °
le *°
2 ) (3 00 liters C 3 H 8 )
\ 1 mole L. 3 rig/
= 150 liters of O 2
(b) If V = the required volume of air (also a gas) that is 21 0% O 2 , you can see that
0 210V liters of air must provide 15 0 liters of O 2 That is,
0 210V = 15 0 liters O 2
V = ;prr7r =714 liters of air