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3 years ago

How many moles of C2H2 are needed to react completely with 84.0 mol O2?

Chemistry

2 answers:

docker41 [41]3 years ago

5 0

Answer: 33.6 moles

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Combustion of ethylene is represented by following balanced equation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to stoichiometry:

5 moles of $O_2$ react completely with = 2 moles of $C_2H_2$

Thus 84.0 moles of $O_2$ react completely with = $\frac{2}{5}\times 84.0=33.6$ moles of $C_2H_2$

Thus 33.6 moles of $C_2H_2$ are needed to react completely with 84.0 mol $O_2$

Airida [17]3 years ago

3 0

33.6 moles are needed to completely react with 84.0 moles of O2

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C2h4 3 o2 2 co2 2 h2o

Marat540 [252]

Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2

O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving

0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2

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3 years ago

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

nevsk [136]

Answer:

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

$Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }$

The given reaction involves the decomposition of H2O into H2 and O2

$2H_{2}O\rightleftharpoons 2H_{2} + O_{2}$

The equilibrium constant is expressed as :

$Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Since Keq = 5.31*10^-10

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

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well hello there ms gabby I might not know the answer but you could text me on Ig or sc

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