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professor190 [17]
3 years ago
12

CH4 (g) + O2 (g) → CO2 (g) + H2O (g) to. If 28.6 g of CH4 is reacted with 57.6 g of O2, calculate the number of grams of CO2 pro

duced. b. If you actually get 32.1 g of CO2, calculate the percent yield. c. Calculate the number of moles of excess reagent remaining at the end of the reaction.
Chemistry
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

See explanation

Explanation:

First we must obtain the limiting reactant. The equation of the reaction is;

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Number of moles of CH4= 28.6 g/16g/mol = 1.8 moles

Since the reaction is 1:1, 1.8 moles of CO2 was produced

Number of moles of O2 = 57.6 g/32 g/mol = 1.8 moles

2 moles of O2 produced 1 mole of CO2

1.8 moles of O2 produced 1.8 × 1/2 = 0.9 moles of CO2

Hence O2 is the limiting reactant

Mass of CO2 produced = 0.9 moles × 44 g/mol = 39.6 g

%yield = 32.1g/39.6 g × 100

%yield = 81.1%

According to the reaction equation;

2moles of O2 reacts with 1 mole of CH4

1.8 moles of O2 reacts with 1.8 × 1/2 =0.9 moles of CH4

Number of moles of CH4 left = 1.8 moles - 0.9 moles

Number of moles of CH4 left = 0.9 moles

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Answer:

the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Which of the following gases would be most likely to experience ideal behavior at high pressures?
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3 years ago
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The mass of a silver bracelet is 2.5 gram, it occupies a volume of 48cm3, what is its density?​
NNADVOKAT [17]

Answer:

The answer is

<h2>0.052 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 2.5 g

volume = 48 cm³

The density is

density =  \frac{2.5}{48}   \\  = 0.052083333...

We have the final answer as

<h3>0.052 g/cm³</h3>

Hope this helps you

3 0
3 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0
3 years ago
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
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actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

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Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

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Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
3 years ago
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