Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
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Answer:
The answer is 4.28 moles
Explanation:
This is super easy okay, you won't forget this!
Basically mole ratios, we're just looking at the coefficients in front of the compounds, multiplying them, and dividing them as we see fit.
In this example, you can see how you need 2 moles of lithium bromide (LiBr) for the reaction, and 2 moles of lithium chloride (LiCl) will be produced.
Basically, the <u>molar ratio</u> is when you divide numbers and see how much of this do I have for that (if that makes sense).
So if you were to divide the 2 moles of LiBr / 2 moles of LiCl = 1. So we know that the mole ratio for LiBr to LiCl is 1:1 or 2:2, either or, it's the same thing.
SO THE BIG IDEA, if we have 4.28 moles of lithium bromide reacting, we should also have 4.28 moles of lithium chloride produced, BECAUSE the <u>mole ratio</u> is 1:1.
I hope this makes sense please tell me if it doesn't, I will try my best to explain a little more.
