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Rom4ik [11]
3 years ago
9

When methane is burned with oxygen, the products are carbon dioxide and water, If you produce 20 grams of water and 10 grams of

carbon dioxide from 15 grams of oxygen, how many grams of methane were needed for the reaction?​
Chemistry
2 answers:
serious [3.7K]3 years ago
5 0

Answer:

4 g

Explanation:

First you need to write a balanced chemical equation. You are given thatmethane is burned, meaning a combustion reaction in which carbon dioxide and water are released.

Unbalanced: CH4 + O2 ---> CO2 + H2O

Balanced: CH4 + 2O2 ---> CO2 + 2H2O

Givens:

X grams CH4 (Molecular mass 16.0 grams)

9 grams H2O (Molecular mass 18.0 grams)

11 grams CO2 (Molecular mass 44.0 grams)

Mole ratio: 1:2:1:2 (CH4:O2:CO2:H2O)

Then you need to find which of the reactants are the limiting reactant and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.

11 g CO2 / (44.0 g) = 0.25 moles CO2

9 g H2O / (18.0 g) = 0.5 moles

n of CH4 = n of CO2 = n of H2O /2 = 0.25 moles

m of CH4 = n* Mw = 0.25 * 16.0 = 4 g

scZoUnD [109]3 years ago
3 0
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
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What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?
zubka84 [21]
Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g 
H: 16.3% = 16.3 g 

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

C:  \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol

H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

C:  \dfrac{6.975}{6.975} = 1

H:  \dfrac{16.3}{6.975} \approx 2.3

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
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\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark

I hope this helps. =)
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