C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(g)
If 9.2g of C2H5OH(l) burns completely in the presence of excess O2(g) according to the equation, how many grams of CO2(g) are produced?
A:0.40g
B:8.8g
C:9.2g
D:18g

Answer:

D. 18g

Explanation:

To get the answer, we will use an equation to convert the grams of C2H5OH to moles of C2H5OH, then I will convert the moles of C2H5OH to moles of CO2, and then I will convert the moles of CO2 to grams of CO2.

9.2g C2H5OH / 46.07g C2H5OH = 0.199 mol C2H5OH

0.199 mol C2H5OH (2 mol CO2 / 1 mol C2H5OH) = 0.398 mol CO2

0.398 mol CO2 (44.01g CO2 / 1 mol CO2) = 17.515g CO2

The answer closest to our answer is answer choice D. 18g

Therefore, the answer is D

In normal temperatures, carbon dioxide is a colorless, non-flammable gas, its calculated value is "18 gram".

Equation:

[tex]\to \bold{C_2H_5OH\ (l)+3O_2\ (g) \to 2CO_2\ (g)+3H_2O\ (g)}[/tex]

In the above-given scenario, We'll utilize an equation that converts grams of ethanol to moles of ethanol, then converts moles of ethanol to moles of carbon dioxide, and finally, transforms the moles of carbon dioxide into grams of carbon dioxide to get the result.

Following are the calculation of the conversion of carbon dioxide:

[tex]\to \frac{9.2\ g\ C_2H_5OH }{46.07\ g\ C_2H_5OH} = 0.199\ mol C_2H_5OH\\\\\to 0.199\ mol\ C_2H_5OH \ (\frac{2\ mol\ CO_2}{ 1\ mol\ C_2H_5OH}) = 0.398\ mol\ CO_2\\\\\to 0.398\ mol \ CO_2 \ (\frac{44.01\ g\ CO_2}{1 \ mol\ CO_2}) = 17.515\ g\ CO_2 \approx 18\ g \ CO_2\\\\[/tex]

Therefore, the answer is "Option D".

Find out more about the CO2 here:

brainly.com/question/867804

Read the temp what is it?
17.3 or 17.25??