In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
Answer:
The answer to your question is: letter D
Explanation:
In a combustion reaction, the reactants are always a molecule with Carbon that reacts with oxygen and the products are carbon dioxide and water.
According to the explanation, the only possible solution is:
a) C₆H₁₂O₂(l) ⇒ 6 C(s) + 6 H₂(g) + O₂(g)
b) Mg(s) + C₆H₁₂O₂(l) ⇒ MgC₆H₁₂O₂(aq)
c) 6 C(s) + 6 H₂(g) + O₂(g) ⇒ C₆H₁₂O₂(l)
d) C₆H₁₂O₂(l) + 8 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(g)
e) None of the above represent the combustion of C₆H₁₂O₂.
A.) AlO is the correct formula
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