2 K + Cl2 ===> 2 KCl
from 1.00g of Cl2 and excess K?
how do you solve this equation?
2007-01-21 13:56:37 · 2 answers · asked by J.r. 3 in Science & Mathematics ➔ Chemistry
This question is unclear. What are you tryng to do?
Do want to know how much KCl you produce?
If you have an excess of K, that means Cl2 is the limiting reactant.
Calculate the number of moles of Cl2 in 1g
the molecular mass of Cl2 is 70.9
you have 1 gram, so thats 1/70.9 = 1.41x10^-2 moles
the mole ratio is 2 moles of KCl per 1 mole of Cl2, so you produce
(1.41x10^-2)x2 = 2.82x10^-2 moles of KCl
the molecular mass of KCl is 74.5 so you produce
74.5(2.82x10^-2) = 2.1 g
2007-01-21 14:04:33 · answer #1 · answered by Anonymous · 1⤊ 0⤋
I'm not sure what is meant by "solving" this equation.
Cl has a molecular weight of 35.45 grams/mol
thus Cl2 has a MW of 70.90
1 gram Cl2/ (70.90 grams/mole Cl2) => .014 moles Cl2
From the equation you can see that you make 2 moles of KCl for each mole of Cl2. That means .028 moles KCl.
That is how much potassium chloride (KCl) would be produced if all the Cl2 were completely converted.
I don't know if that "solves" the equation or not.
2007-01-21 22:05:40 · answer #2 · answered by enginerd 6 · 0⤊ 1⤋