Mass-Volume Problems

Question

Sugar cane plants convert carbon dioxide (CO2) and water (H20) to sucrose (C12H22O11) and oxygen (O2) in the presence of sunlight according to the following reaction:

12CO2(g) + 11H2O(l) --> C12H22O11(s) + 12O2(g)

1. How many grams of sucrse (C12H22O11) is produced from 224 cubic decimeters of carbon dioxide (CO2) at STP?

2. How many cubic decimeters of carbon dioxide (CO2) at STP is needed to produce 5.00 pounds of sugar? (1kg=2.20 lbs.)

3. what mass of water would be needed to combine with 200 cubic decimeters of CO2 at STP?

One of the steps in the production of iron utilizes the following chemical raction:

3CO(g) + Fe2O3(s) --> 2Fe(s) + 3CO2(g)

4. What mass of Fe2O3 would react with 50 cubic decimeters of C) at STP?

5. What volume of carbon dioxide (CO2) at STP is produced from 1000 grams of Fe2O3?

6. What mass of iron (Fe) is produced when 300 cubic centimeters of C02 is produced at STP?

Answer 1

12 CO2(g) + 11 H20(l) -> C12H22O11(S) + 12 O2(g)

( 1 )

g C12H22O11 = ?

Given: 224L CO2 @ STP (1 cubic decimeter = 1 Liter)

a) find moles of CO2 in reaction

PV=nRT

P = 1 atm

V = 224 L

R = 0.082057 atm * L / mol * K

T = 273.15 K

n = ?

n = PV / RT = (1 * 224) / (0.082057 * 273.15) = 9.994 mol CO2

b) convert moles CO2 to moles C12H22O11

9.994 mol CO2 * (1 mol C12H22O11 / 12 mol CO2) = 0.8328 mol C12H22O11

c) convert moles C12H22O11 to grams C12H22O11

0.8328 mol C12H22O11 * ( 12*12.010 + 22*1.0079 + 11*15.999) g C12H22O11 / 1 mol C12H22O11 = 285.1 g C12H22O11

[ 285.1 g C12H22O11 ]

( 2 )

L CO2 @ STP = ? (1 cubic decimeter = 1 Liter)

Given: 5.00 lb C12H22O11 (1 kg = 2.20 lb)

a) find moles of C12H22O11 in reaction

5.00 lb C12H22O11 * ( 1000 g / 2.20 lb) * ( 1 mol C12H22O11 / 342.28 g C12H22O11) = 6.640 mol C12H22O11

b) convert moles C12H22O11 to moles CO2

6.640 mol C12H22O11 * (12 mol CO2 / 1 mol C12H22O11) = 79.68 mol CO2

c) Ideal Gas Law to find Volume in Liters

PV=nRT

P = 1 atm

n = 79.68 mol CO2

R = 0.082057 atm * L / mol * K

T = 273.15 K

V = ?

V = nRT / P = 79.68 * 0.082057 * 273.15 /  1 = 1785.93 L CO2

[ 1785.93 L CO2 ]

( 3 )

g H2O = ?

Given: 200L CO2 @ STP (1 cubic decimeter = 1 Liter)

a) find moles of CO2 in reaction

PV=nRT

P = 1 atm

V = 220 L

R = 0.082057 atm * L / mol * K

T = 273.15 K

n = ?

n = PV / RT = (1 * 220) / (0.082057 * 273.15) = 9.815 mol CO2

b) convert moles CO2 to moles H2O

9.815 mol CO2 * ( 11 mol H2O / 12 mol CO2 ) = 8.997 mol H20

c) convert moles H2O to grams H2O

8.997 mol H20 * (2 * 1.0079 + 15.999) g H20 / 1 mol H20 = 162.086 g H2O

[ 162.086 g H2O ]



3 CO(g) + Fe2O3(s) -> 2 Fe(s) + 3 CO2(g)

( 4 )

g Fe2O3 = ?

Given: 50L CO @ STP (1 cubic decimeter = 1 Liter)

a) find moles of CO in reaction

PV=nRT

P = 1 atm

V = 50 L

R = 0.082057 atm * L / mol * K

T = 273.15 K

n = ?

n = PV / RT = (1 * 50) / (0.082057 * 273.15) = 2.231 mol CO

b) convert moles CO to moles Fe2O3

2.231 mol CO * ( 1 mol Fe2O3 / 3 mol CO ) = 0.7436 mol Fe2O3

c) covnert moles Fe2O3 to grams Fe2O3

0.7436 mol Fe2O3 * ( 2 * 55.845 + 3 * 15.999 ) g Fe2O3 / 1 mol Fe2O3 = 118.7 g Fe2O3

[ 118.7 g Fe2O3 ]

( 5 )

L CO2 @ STP = ? (1 cubic decimeter = 1 Liter)

Given: 1000g Fe2O3

a) find moles of Fe2O3 in reaction

1000 g Fe2O3 * ( 1 mol Fe2O3 / 159.687 g Fe2O3 ) = 6.262 mol Fe2O3

b) convert moles Fe2O3 to moles CO2

6.262 mol Fe2O3 * (3 mol CO2 / 1 mol 6.262 mol Fe2O3) = 18.787 mol CO2

c) Ideal Gas Law to find Volume in Liters

PV=nRT

P = 1 atm

n = 18.787 mol CO2

R = 0.082057 atm * L / mol * K

T = 273.15 K

V = ?

V = nRT / P = 18.787 * 0.082057 * 273.15 /  1 = 421.1 L CO2

[ 421.1 L CO2 ]

( 6 )

g Fe = ?

Given: 300L CO2 @ STP (1 cubic decimeter = 1 Liter)

a) find moles of CO2 in reaction

PV=nRT

P = 1 atm

V = 300 L

R = 0.082057 atm * L / mol * K

T = 273.15 K

n = ?

n = PV / RT = (1 * 300) / (0.082057 * 273.15) = 13.385 mol CO2

b) convert moles CO2 to moles Fe

13.385 mol CO2 * ( 2 mol Fe / 3 mol CO2 ) = 8.9230 mol Fe

c) covnert moles Fe2O3 to grams Fe2O3

8.9230 mol Fe * ( 55.845 g Fe / 1 mol Fe ) = 498.3 g Fe

[ 498.3 g Fe ]