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bogdanovich [222]
3 years ago
11

In an experiment, 8.50 g of methane, CH4, was reacted with 15.9 g of oxygen gas to produce carbon dioxide and water. Determine t

he percentage yield if 9.77 g of carbon dioxide was obtained in the lab.
Chemistry
1 answer:
exis [7]3 years ago
7 0

Answer:

89.3 %

Explanation:

M(CH4) = 12+ 4*1 = 16 g/mol

M(O2) = 2*16 = 32 g/mol

M(CO2) = 12 + 2*16 = 44 g/mol

8.50 g * 1 mol/16 g = 0.5313 mol CH4

15.9 g * 1 mol/32 g = 0.4969 mol O2

9.77 g * 1 mol/44 g = 0.2220 mol CO2

1)                                  CH4        + 2O2 -----> CO2 + 2H2O

from reaction             1 mol          2 mol

given                       0.5313 mol   (0.4969 mol)

1 mol CH4              --- 2 mol O2

0.5313 mol  CH4  --- x mol O2

x= 2*0.5313 = 1.0626 mol O2

We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.

2)                               CH4        + 2O2 -----> CO2 + 2H2O

from reaction                             2 mol        1 mol

given                                      0.4969 mol   x mol

x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield

3)

Practical yield CO2 = 0.2220 mol

Theoretical yield CO2 = 0.2485 mol

% yield = (0.2220/0.2485)*100% = 89.3 %

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A mixture of c2h2 and ch4 has a total mass of 230.9 g. when this mixture reacts completely with excess oxygen, the co2 and h2o p
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Interesting problem. Thanks for posting.

C2H2 + (3/2)02 ====>  H2O + 2CO2
CH4 +  2O2 =====> 2H2O + CO2

The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16

The number of moles of C2H2 = x
The number of moles of  CH4 = y
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For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18

x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water. 

Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
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From CH4 we get 1*y*44 grams of CO2
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Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y =  972.7

Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7

I'm not going to go through the math unless you request me to do so. 
x = 8.03 moles
y = 1.38 moles 

The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
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