Question

The two balanced equations (1) and (2) are for reactions in which gaseous carbon dioxide is produced from the combustion of (1) solid carbon and (2) gaseous carbon monoxide.

1. C(s) + O2(g) CO2(g) + 94.0 kcal



2. CO(g) + ½O2(g) CO2(g) + 67.6 kcal

When 112 grams of carbon monoxide are consumed according to equation (2), which of the following occurs? (atomic weights: C = 12.0 g/mol, O = 16.0 g/mol)

Answer 1

The question says: which of the following occurs?
Where are the following?

Answer 2

1.0 mole of carbon dioxide is produced.
67.6 kcal of heat are generated.
2.00 moles of oxygen are consumed.
0.25 mole of carbon dioxide is produced.
0.50 mole of oxygen is consumed.

Here is the following

Answer 3

Oke so we have to use reaction 2

\(\Large\sf CO(g)+\frac{1}{2}O_2(g)\to CO_2(g)\)
Let's get rid of the fractionby multiplying the whole equation with 2.

\(\Large\sf 2CO(g)+O_2(g)\to 2CO_2(g)\)

112 grams of CO is used in the reaction
Convert that to moles. Do you know how to do that?

Answer 4

Are you there @Miguel27525 ?

Answer 5

No in dont know how...

Answer 6

You divide 112 by the molar mass of carbon monoxide

Answer 7

Umm so the molar mass of carbon monoxide is 28.01 right?

Answer 8

C = 12.011
O = 15.999
CO=12.011+15.999=28.01 g/mol

So yea :)

Answer 9

So you divide 112/28.01?

Answer 10

Yes, then you know how many moles of CO you have for the reaction

Answer 11

0.25 moles?

Answer 12

28.01 g/mol means 1 mol = 28.01 gram
You have 112 gram, how can your answer be less than 1 mol? :P

Answer 13

sorry sorry sorry :P it's 2.00 right?

Answer 14

\(\large\dfrac{112}{28.01}=3.998~mol\)

Answer 15

Ohhh

Answer 16

for 2 mol CO, we get 2 mol CO2

So when we have 3.998 mol CO, we get 3.998 mol CO2

Only thing you have to do is convert 3.998 mol CO2 to grams.
Multiply with the molar mass of CO2

Answer 17

Thank you :D

Answer 18

You're welcome :)