How many moles is 573.28 g of AgCl?

4 moles

Explanation:

First, let's find the molar mass of AgCl. The molar mass of Ag is 107.87 and the molar mass of chlorine is 35.45 so the total mass is 107.87 + 35.45 = 143.32 grams per mole. Now all we need to calculate is 573.28 / 143.32 = 4 moles.

4 mols

Explanation:

Step 1: Find molar mass of AgCl

Ag - 107.8682

Cl - 35.45

Ag + Cl = 143.318 g/mol

Step 2: Conversion

We should get 4.00005, rounded to 4 mols as our answer.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

answer: c.

explanation: