# How many mole(s) of Oxygen is/are required to combust 1.5moles of CH4 ?

Update:

My answer comes out to 3moles of oxygen. Is this what you get?

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write the balanced equation:

CH4 + 2 O2 --> CO2 + 2 H2O

get the mole ration of CH4 to O2:

1 mol CH4 : 2 mol O2

1.5 mole CH4 X 2 = 3.0 mole O2 required

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• Anonymous

CH4 + 2O2 = CO2 + 2H2O

moles of CH4 given is 1.5

moles of O2 will be 1.5x2=3moles

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• Anonymous

first of all balance the equation. the balanced eq for CH4 conbustion is

CH4 + 2 O2 --> CO2 + 2 H2O

1.5 moles of CH4 can react with on 2 moles of O2

So

No of moles of O2 to burn 1.5 moles of CH4=2*1.5(no of moles of CH4)

=3 moles of O2

Nawaz Bajwa. Pakistan

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• CH4+2O2--->CO2+2H2O

you will need 3 moles of O2 to completely combust 1.5 moles of CH4

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• CH4 + 2O2 ------> CO2 + 2H2O

CH4 : O2

1 : 2 ....... from the the reaction

1.5 : x

cross mutiply; x=1.5 X2 ...... so, x=3 moles

yes,it is 3 moles

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• Read the following answer I wrote up, paying special attention to the last part on "Gas Problems":

Too solve these types of problems, use the following general steps:

[I will use the general word equation: calcium hydroxide reacts with lithium bromide to produce calcium bromide and lithium hydroxide; as an example, throughout each step, to aid in your learning]

(1) Write the correct chemical formula for each substance in the reaction. Be certain each chemical equation is correct! (ex: calcium hydroxide: correct formula = Ca(OH)2; NOT CaOH.)

(2) Write out the chemical equation; ex: Ca(OH)2 + LiBr gives CaBr2 + LiOH

(3) Balance this chemical equation; ex: Ca(OH)2 + 2LiBr gives CaBr2 + 2LiOH

Note: this balanced equation is TELLING you that one mole of calcium hydroxide is reacting with 2 moles of lithium bromide to produce (gives) one mole of calcium bromide and 2 moles of lithium hydroxide. It is TELLING you the relative number of moles of each substance.

(4) Determine the MOLE RATIO of the substances involved; ex: if the question says you have 1.0 moles of calcium hydroxide and it reacts with an excess of lithium bromide, and asks how many moles of lithium hydroxide are produced, the answer would be 2.0 moles of lithium hydroxide. In other words, the MOLE RATIO of calcium hydroxide to lithium hydroxide is 1 to 2, also written as a “1:2 ratio“. ex: same question but suppose you have 3 moles calcium hydroxide. Because the mole ratio is 1:2, you would produce 6 moles of lithium hydroxide. A ratio of 3:6 = a ratio of 1:2; (you just reduced it to simpler terms).

Stop here if you are simply given the number of moles of product you have, and asking how many moles of some different substance are produced in a chemical reaction.

Same type of problems, but working with grams instead of moles:

you simply have to do a couple of extra steps:

I will use the same reaction as above, but will change the question to this:

“If you have 255 grams of calcium hydroxide, how many grams of lithium hydroxide will be produced, (assuming you have an excess of lithium bromide)”.

[Note: unless the question states otherwise, assume you have an excess of the other reactant, thus giving the maximum yield of product; see “Limiting Reagents” below to solve those types of problems].

To solve, complete the first 4 steps above (we’ve just done that):

(5) Convert grams of substance into moles of substance!!! This is essential in solving these types of problems. THE KEY STEP FOR ALL PROBLEMS OF THIS TYPE IS TO ALWAYS CONVERT EVERYTHING INTO MOLES FIRST!!! The general formula’s are as follows:

Moles = grams / molecule weight

Or you can use basic algebra and say the following:

Grams = Moles X molecular weight

[Note: “molecular weight” is used to describe molecules; “formula weight” is used to describe ionic compounds; and “atomic weight” is used to describe elements; ALL of them are the sum of all the atomic weights of the elements in the substance. In the case of a single element, alone by itself, this refers to the atomic weight of that element. Note that the 7 elements: hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine only exist as “diatomic” elements, when they are ALONE by themselves (not combined with any other element(s)). Thus H2, is the correct way to write the chemical formula for hydrogen when it is all alone, by itself. And it is a molecule, and it’s molecular weight would be equal to 2.0. This is the same for the all diatomic molecules: the molecular weight will equal double the atomic weight for that particular diatomic element (because there are TWO atoms, not just one, in these diatomic molecules). This is ONLY true for these seven elements, when they appear ALONE in a chemical equation].

{The example I have shown here does NOT contain any diatomic elements, but if it did, the problems are still solved in a similar manner}.

Ex: if we have 255 g of calcium hydroxide, convert this to moles

Moles Ca(OH)2 = 255 g of Ca(OH)2 / 78.1 g/mol Ca(OH)2

= 3.27 mol Ca(OH)2

(6) According to the mole ratio, determined in step 4 above, you have a mole ratio of 1 mol Ca(OH)2 gives 2 moles of LiOH; in other words, you have a 1:2 mole ratio.

Ex: thus, 3.27 moles Ca(OH)2 will give 6.54 moles of LiOH [2 x 3.27 = 6.54]

(7) To convert 6.54 moles of LiOH into grams, use the second formula given above:

Ex: Grams = moles X molecular weight

Grams LiOH = 6.54 moles LiOH X 23.9 grams/mole LiOH

= 156.3 grams LiOH

LIMITING REAGENTS: Same type of problem, except this time you have a Limited amount of reagent:

I will use the same chemical reaction as above, but will change the question to this:

“If you have 255 grams of calcium hydroxide, and 52.0 grams of lithium bromide, how many grams of lithium hydroxide will be produced?”

To solve, complete the first 4 steps above (we’ve just done that).

(5a) We’ve already calculated that we have 3.27 mole of Ca(OH)2 We will use that later, so save that answer. Now, calculate the number of moles of lithium bromide you have if you have 52.0 grams of lithium bromide:

Moles LiBr = 52.0 g LiBr / 86.8 g/mol LiBr

= 0.60 mole of LiBr

(6A) Notice that the mole ratio of the REACTANTS, calcium hydroxide and lithium bromide are in a 1:2 ratio. This is TELLING you that you will need 2 X 0.600 moles LiBr = 1.20 moles of LiBr to completely react the 3.27 mole of Ca(OH) 2 . But you don’t HAVE 1.27 moles, you only have 0.60 moles. Thus lithium bromide is the LIMITING REACTANT. Furthermore, since you only have 0.60 mole LiBr, you will ONLY use up 0.30 mole of Ca(OH)2 . [remember, the mole ratio is 1:2]

(7) To determine how much LiOH will be made, simply follow step 6 above, but use 0.30 mole Ca(OH)2 .

Ex: using the mole ratio of 1:2, as you did in step 6, you will see that you will produce 0.60 mole of LiOH.

(8) Convert 0.60 mole of LiOH into grams LiOH, in the same manner as you did in Step 7:

Grams = 0.60 moles LiOH X 23.0 grams/mole LiOH

= 13.8 grams LiOH will be produced.

And LiBr is the limiting reactant (sometimes called “reagent”).

GAS PROBLEMS: the steps are very similar. Follow steps 1 through 4. But beginning with Step 5, use the FACT, that for gases:

1 mole of any gas (at STP) occupies a space of 22.4 liters

In other words, use the conversion factor:

Moles = liters / 22.4 liters per mole

Or Liters = Moles X 22.4 liters/mole

Use either of the these 2 equations to convert back and forth from liters to moles.

[Note: This is ONLY true for Ideal gases, at standard temperature and pressure (STP) conditions.Unless the question mentions otherwise, assume they are using ideal gases at STP.]

Good luck and have fun!

Your friendly retired PUBLIC school chemistry teacher.

Source(s): rct
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