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3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

Chemistry

1 answer:

luda_lava [24]3 years ago

6 0

the balanced equation is CH4 + 2O2 -> CO2 + 2H20

moles = mass ÷ mr

moles of methane = 10 ÷ 16

= 10/16

= 5/8 moles

the moles of carbon dioxide are the same as methane because the ratio is 1:1

mass = moles × Mr

mass of carbon dioxide = 5/8 × 44

= 220 ÷ 8

= 27.5 grams

to find out whether it obeys the law of conservation of mass, you must find the mass of the reactants and the mass of the products. if they are the same, it obeys the law of conservation of mass. otherwise it doesnt.

the mass of oxygen is 32 × 5/8

= 160/8

= 20g

we already have the mass of methane (10g) so add it on

mass of reactants = 20 + 10 = 30g

find mass of water: 5/8 × 18

= 90/8

= 11.25g

mass of products = 11.25 + 27.5 = 38.75g

the mass of reactants vs products are not the same so it does not obey the law of conservation of mass

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Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive

lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene. The constant Kc for the above reaction is,

Kc = $\frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}$

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PV = nRT

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We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,

$[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}$

$[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}$

So, the molar concentration of ethene (C₂H₄) will be,

$[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}$

Then, according to the law of ideal gases,

$P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L} x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm$

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

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