Your answer is D. You're Welcome
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl

(ii) From O₂

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used

(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
Answer:

Explanation:
Hello,
In this case, we use the ideal gas equation to compute the volume as shown below:

Nonetheless we are given mass, for that reason we must compute the moles of gaseous fluorine (molar mass: 38 g/mol) as shown below:

Thus, we compute the volume with the proper ideal gas constant, R:

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