168.96 g of carbon dioxide (CO₂)
Explanation:
The chemical reaction representing the combustion of acetylene:
2 C₂H₂ (g) + 5 O₂ (g)→ 4 CO₂ (g) + 2 H₂O (g)
number of moles = mass / molecular weight
number of moles of acetylene (C₂H₂) = 50 / 26 = 1.92 moles
Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:
if 2 moles of acetylene (C₂H₂) produces 4 moles of carbon dioxide (CO₂)
then 1.92 moles of acetylene (C₂H₂) produces X moles of carbon dioxide (CO₂)
X = (1.92 × 4) / 2 = 3.84 moles of carbon dioxide (CO₂)
mass = number of moles × molecular weight
mass of carbon dioxide (CO₂) = 3.84 × 44 = 168.96 g
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combustion of hydrocarbons
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Answer:
The composition of air this is because it vmade up of oxygen, nitrogen, Nobel gages and Carbon dioxide
Answer: The mass percentage of
is 5.86%
Explanation:
To calculate the mass percentage of
in the sample it is necessary to know the mass of the solute (
in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the
precipitate. We can establish a relation between the mass of
and
using the stoichiometry of the compounds:

Since for every mole of Tl in
there are two moles of Tl in
, we have:

Using the molar mass of
we have:

Finally, we can use the mass percentage formula:
