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Gnesinka [82]
3 years ago
14

LOOK AT THE IMAGE ABOVE CAN SOMEONE PLEASE DO IT WITH FULL STEPS PLEASE I NEED IT TODAY PLEASE PLEASE I WILL MARK YOU BRAINLIST

Chemistry
1 answer:
Bogdan [553]3 years ago
3 0

answer:

IMAGE IS SUPER UNCLEAR SORRY LUV <3

explanation:

your phone must me from 2009 with that trash camera quality sweetie i suggest you get a new one... oh wait you can't you're broke :(

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

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B. The collisions release heat, which results in the heating and subsequent melting, sinking, and rising of materials.

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