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mr Goodwill [35]
3 years ago
15

For the reaction, 2A + B → 3C, rate of disappearance of A is 0.3 M/s in a certain time interval.What is the average rate of appe

arance of C and what is the average rate of the reaction in that time interval?
A) rate of appearance of C = 0.45 M/s
rate of the reaction = 0.6 M/s
B) rate of appearance of C = 0.2 M/s
rate of the reaction = 0.6 M/s
C) rate of appearance of C = 0.2 M/s
rate of the reaction = 0.15 M/s
D) rate of appearance of C = 0.45 M/s
rate of the reaction = 0.3 M/s
E) rate of appearance of C = 0.45 M/s
rate of the reaction = 0.15 M/s
Chemistry
1 answer:
andrezito [222]3 years ago
7 0

Answer:

E) rate of appearance of C = 0.45 M/s

rate of the reaction = 0.15 M/s

Explanation:

2A + B → 3C

Writing rate law for the reaction:

<u>Rate of reaction</u> = -\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3}\frac{d[C]}{dt} → equation 1

Given that the rate of disappearance of A is 0.3 M/s

⇒ - \frac{d[A]}{dt} = 0.3 M/s

⇒Rate of reaction = - \frac{1}{2}\frac{d[A]}{dt} = \frac{1}{2}×0.3 M/s

⇒<u>Rate of reaction = 0.15 M/s</u>

From equation 1, \frac{d[C]}{dt} = - \frac{3}{2}\frac{d[A]}{dt} = \frac{3}{2}×0.3 M/s

⇒\frac{d[C]}{dt} = 0.45 M/s

or <u>the rate of appearance of C = 0.45 M/s</u>

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Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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Explanation:

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Option D is false because losing one or more electron will turn an atom to a cations.

7 0
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