Pentane, C5H12, will react as shown below:

C5H12(l) + O2 --- CO2 +_____ H2O

H20(g)

1. Balance the equation above by putting coefficients into each blank space.

1C5H12+ 8O2----- 5CO2+ 6 H2O (this is what I got)

2. How many atoms of oxygen are in 144 grams of pentane?

step 1: 144g of C5H12(1mole/72.17)= 1.995moles of C5H12
step 2: 1.995moles of C5H12(8moles of O2/1mole of C5H12)= 15.962 moles of O2
step 3: 15.962 moles of O2(6.0210^23 atoms of O2/1mole of O2)= 9.60910^24 atoms of O2 (I don't know if I did this right)

3. How many atoms of oxygen are in 192 grams of carbon dioxide gas?

step 1: 192 grams of CO2 (1mole of CO2/44.01grams of CO2)= 4.363 moles of CO2
step two: 4.36moles of CO2* ( 1 mole of O2/1mole of CO2)=
step 3: 4.363* (6.0210^23) = 2.62710^24 atoms of O2 (I don't know if this is correct.

4. A) CIRCLE ONE: True or False:

reactant you begin with fewer grams of is your limiting reactant.

B) If 32 g of oxygen gas and 32 g of pentane are combined, how much water will be produced, in moles? (Use your reaction from the top of the first page)

NEED HELP ASAP THANK YOU!!!

Question

ampritchard ampritchard

01-11-2014
Chemistry

Answered

Pentane, C5H12, will react as shown below:

C5H12(l) + O2 --- CO2 +_____ H2O

H20(g)

1. Balance the equation above by putting coefficients into each blank space.

1C5H12+ 8O2----- 5CO2+ 6 H2O (this is what I got)

2. How many atoms of oxygen are in 144 grams of pentane?

step 1: 144g of C5H12(1mole/72.17)= 1.995moles of C5H12
step 2: 1.995moles of C5H12(8moles of O2/1mole of C5H12)= 15.962 moles of O2
step 3: 15.962 moles of O2(6.0210^23 atoms of O2/1mole of O2)= 9.60910^24 atoms of O2 (I don't know if I did this right)

3. How many atoms of oxygen are in 192 grams of carbon dioxide gas?

step 1: 192 grams of CO2 (1mole of CO2/44.01grams of CO2)= 4.363 moles of CO2
step two: 4.36moles of CO2* ( 1 mole of O2/1mole of CO2)=
step 3: 4.363* (6.0210^23) = 2.62710^24 atoms of O2 (I don't know if this is correct.

4. A) CIRCLE ONE: True or False:

reactant you begin with fewer grams of is your limiting reactant.

B) If 32 g of oxygen gas and 32 g of pentane are combined, how much water will be produced, in moles? (Use your reaction from the top of the first page)

NEED HELP ASAP THANK YOU!!!

Answer :

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NerdyBirdy NerdyBirdy · Answer 1 · 2014-11-02T07:44:30-05:00

1) That looks right

2) Step 1: Moles=Mass/Mr, so it's 144/72= 2mol
Step 2: Stoichiometry means the ratio of oxygen to pentane is 1:8 so moles of oxygen is moles of pentane times 8. 2*8=16mol of oxygen
Step 3: 16= 9.6*10^24

3) The ratio of oxygen to carbon in CO2 is 1:2
so it's 192/3 = 64
64*2=128g
Moles = Mass/Mr = 128/32 = 4mol = 2.41*10^24 atoms of oxygen
Basically you needed to work out the mass of oxygen using ratios before using a moles equation instead of just working with moles

4) True

5) 1C5H12+ 8O2----- 5CO2+ 6 H2O
32g 32g

Moles of oxygen = Mass/Mr = 32/32= 1mol so you'd need 0.125mol of pentane to react fully
Moles of pentane= Mass/Mr = 32/72=0.4mol so you'd need 3.2mol of oxygen to react fully

This makes oxygen the limiting factor, because you need more of it to react fully with the pentane, so we use oxygen

Stoichiometry means for every 8mol of oxygen there's 5mol of CO2
So, 1/8*5=0.635mol of CO2