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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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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K
swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

Pv =nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 4.3 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

4.3 liter = 4.3\times10^{-3}=4.3\times10^{-3} m^{3}

And initial temperature of balloon, T_{1} = 350 K

Let the final volume of balloon is V_{2}

And as given, final temperature of balloon, T_{2} is 250 K

Now, V_{1} = KT_{1}

4.3\times10^{-3}=K\times350\ (equation\ 1 )

V_{2} = KT_{2}

=K\times250\ (equation 2)

Dividing equation 1 and 2,

\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}

K cancelled by K.

By cross multiplication:

350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\          = \frac{1075\times10^{-3}}{350} \\          =2.87\times10^{-3}m^{3}

Now, convert it into liter with the help of calculation done above,

2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter

Therefore, volume of the gas in the balloon at 250 K will be  2.87 liter.

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