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4vir4ik [10]
3 years ago
9

What is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest whole-number coeffici

ents? C3H6O2(l) + O2(g) → CO2(g) + H2O(l)A) 5 B) 7 C) 1 D) 3
Chemistry
1 answer:
rodikova [14]3 years ago
7 0

Answer:

Option (B) 7

Explanation:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

To know the coefficient of O2 in the above equation, let us balance the equation.

The above equation can be balance as follow:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 3 in front of CO2 as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + H2O(l)

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + 3H2O(l)

There are a total of 4 atoms of O on the left side and a total of 9 atoms on the right side. It can be balance by putting 7/2 in front of O2 as show below:

C3H6O2(l) + 7/2O2(g) → 3CO2(g) + 3H2O(l)

Multiply through by 2

2C3H6O2(l) + 7O2(g) → 6CO2(g) + 6H2O(l)

Now, the equation is balanced.

From the balanced equation above, the coefficient of O2 is 7.

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How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

4 0
3 years ago
Read 2 more answers
If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles
VARVARA [1.3K]

Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

or

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation:

5 0
3 years ago
Determine the total pressure of a mixture of 0.400 mole of He and 0.600 mole of Ne in a 2.00 liter container at 25oC.
harina [27]

Answer:

Total pressure of the mixture is 12.2 atm

Explanation:

Let's apply the Ideal Gases law to solve this

Total pressure . V = Total moles . R . T

Total moles = 0.4 m of He and 0.6 mole of Ne → 1 mol

P . 2L = 1 mol . 0.082 L.atm/mol. K . 298K

P = ( 1 mol . 0.082 L.atm/mol. K . 298K) /2L

P = 12.2 atm

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masha68 [24]
Assuming that the percentage composition is on the basis of weight percentage, therefore the mass of lithium would simply be the product of the two, that is:
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What is the overall order of reaction for this rate law: Rate = k ? zero first second third
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3 0
3 years ago
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