What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?

1 answer:

3 0

2C6H14 + 13O2 ---> 6CO2 +14H2O

Calculating the molar mass of C6H14: M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

Thus, 86.17 g of C6H14 corresponds to 1 mole.

                                  2C6H14 + 13O2 ---> 6CO2 +14H2O
based on the equation        2 mol                            6 mol
according to the question    1 mol                            3 mol

To determine M(CO2): M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
Therefore, 3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Final answer: 132.0 g CO2

You might be interested in

Next we need to determine the mass of oleic acid in the monolayer. The concentration of the oleic acid/benzene solution is 0.02g

Alekssandra [3086]

Response:

$m=1x10^{-6}g$

Clarification:

Hello,

In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

$m=0.02\frac{g}{L}*0.05mL*\frac{1L}{1000mL}\\ \\m=1x10^{-6}g$

Best wishes.

3 0

4 months ago

All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou

VMariaS [2998]

Answer:

$m_{Mg}=30.8mgMg$

Explanation:

Greetings,

According to the provided chemical equation, the production of 31.2 mL of hydrogen allows one to calculate its moles using the ideal gas equation as detailed below:

$n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2$

Since the ratio of hydrogen to magnesium is 1:1, its milligrams are derived through the following proportional factor calculation:

$m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg$

Regards.

3 0

3 months ago

"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

lorasvet [2795]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)