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2 years ago

How many grams of CO₂ will be produced from 7.00 g of C₃H₈ and 98.0 g of O₂?

Chemistry

1 answer:

Alborosie2 years ago

4 0

Answer:

The answer to your question is 21 g of CO₂

Explanation:

Balanced Reaction

C₃H₈ + 6O₂ ⇒ 3CO₂ + 4H₂O

Data

mass of C₃H₈ = 7 g

mass of O₂ = 98 g

Determine the limiting reactant

Molecular mass of C₃H₈ = (12 x3) + (1 x 8) = 36 + 8 = 44 g

Molecular mass of Oxygen = 16 x 12 = 192 g

Theoretical proportion C₃H₈ / O₂ = 44 / 192 = 0.23

Experimental proportion C₃H₈ / O₂ = 7 / 98 = 0.07

As the proportion diminishes in the experiment, the excess reagent is the oxygen and the limiting reactant is propane.

- Calculate the mass of CO₂

44 g of C₃H₈ ---------------- 3(44) of CO₂

7 g of C₃H₈ --------------- x

x = (7 x 3(44)) / 44

x = 924 / 44

x = 21 g of CO₂

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2 years ago

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

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There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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3 years ago