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Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant
3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO
From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂
(b) Mass of NO reacted
(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO