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Alex_Xolod [135]
3 years ago
11

For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

Chemistry
2 answers:
miskamm [114]3 years ago
7 0

Answer:

8.0 mol O₂

Explanation:

Let's consider the complete combustion reaction of C₉H₁₂.

C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O

The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:

0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂

8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.

GenaCL600 [577]3 years ago
6 0

Answer:

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Explanation:

Step 1: Data given

Number of moles C9H12 = 0.67 moles

Step 2: The balanced equation

C9H12 + 12O2 → 9CO2 + 6H2O

Step 3: Calculate moles of O2 required

For 1 mol C9H12 we need 12 moles of O2 to produce 9 moles of CO2 and 6 moles of H2O

For 0.67 moles of C9H12 we need 12 *0.67 = 8.04 moles of O2

To produce 9*0.67 = 6.03 moles of CO2 and 6*0.67 = 4.02 moles H2O

To react with 0.67 moles C9H12 we need 8.04 moles of O2

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B

Explanation:

Ethanol is C2H6O. Carbon diozidex is CO2. Ethanol has hydgrogen, where carbon dioxide does not.

6 0
3 years ago
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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

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7 0
3 years ago
1.If we can see something, it MUST be matter. true or false
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Answer:

False

Explanation:

3 0
3 years ago
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Write and balance the equation for the combination of ammonia gas with solid copper (II) oxide to produce copper metal, nitrogen
Katena32 [7]

Explanation :

The balanced chemical reaction will be,

3CuO(s)+2NH_3(g)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)

By the stiochiometry, 3 moles of solid copper(II)oxide react with 2 moles of ammonia gas to give 3 moles of copper metal, 1 mole of nitrogen gas and 3 moles of liquid water.

The states of matter of each elements and compound is,

Copper(II)oxide is in solid state

Ammonia is in gaseous state

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Nitrogen is in gaseous state

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7 0
3 years ago
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Ivanshal [37]

Answer:

ΔH°r = -1562 kJ

Explanation:

Let's consider the following combustion.

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We can calculate the standard heat of reaction (ΔH°r) using the following expression:

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(i) are the standard heats of formation of reactants and products

The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]

ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]

ΔH°r = -1562 kJ

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