Answer:
The answer to your question is: 17.26% of carbon
Explanation:
Data
CxHy = 0.2121 g
BaCO₃ = 0.6006 g
Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g
Reaction
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Process
1.- Find the amount of carbon in BaCO₃
197 g of BaCO₃ --------------- 12 g of Carbon
0.6006 g ---------------- x
x = (0.6006 x 12) / 197
x = 0.0366 g of carbon
2.- Calculate the percentage of carbon in the organic compound
0.2121 g of organic compound --------------- 100%
0.0366g -------------- x
x = (0.0366 x 100) / 0.2121
x = 17.26%
The percentage of carbon in the sample if 0.6006g barium carbonate was formed is 17.26%.
An organic compound is made up of three elements, carbon, hydrogen, and oxygen.
Given, the sample of the compound 0.2121g
BaCO₃ = 0.6006 g
The molecular mass of BaCO₃ is 197 g
The reaction is
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Step1: Calculating the amount of Carbon in BaCO₃
Now, 197 g of BaCO₃ is = 12 g of carbon
So, 0.6006 g of BaCO₃ will be
[tex] \rm \dfrac{0.6006 \times 12}{197} = 0.0366 g[/tex]
step2: Calculating the percentage of carbon
0.2121 g of organic compound = 100%
0.0366g will be
[tex] \rm \dfrac{ (0.0366 \timess 100)}{0.2121} = 17.26%[/tex]
Thus, the percentage of carbon in the sample is 17.26 g.
Learn more about the organic compounds, here:
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