how many grams of sliver chloride are produced from 5 grams of silver nitrate reacting with an excess of Ba?

Question

2AgNO3+ BaCL2-------- 2AgCl+Ba(NO3)2
how much of barium chloride is necessary to react with the silver nitrate in this problem

Answer 1

All of the different questions you've asked are dealt with in the same way. Convert the mass you're beginning with into moles. Convert moles of that to moles of something else, and finally convert moles of that substance back into mass.

It's important that you recognize that all of these are solved in the same manner.

Answer 2

ok. you have your balanced equation. next convert fifty two g of BaCl2 into moles moles = fifty two g / 208.23 g/mol = 0.2497 moles Now use the mole ratio as given via the equation between BaCl2 and AgCl. are you able to be certain that for each mole of BaCl2 you have you will produce two times that many moles of AgCl? So moles of AgCl produced would be 0.2497 moles x 2 = 0.4994 moles Now convert that many moles of AgCl right into a mass mass = moles x molar mass ok?

Answer 3

Simplest way:

For every 2 AgNO3 you get 2 AgCL.

The atomic weight of AgNO3 = 169.873.
The atomic weight of AgCl = 143.321

So 143.321/169.873 * 5 g = 4.218 g

The atomic weight of BaCl2 = 208.236.

(208.236/(169.873)/2 * 5 g = 3.065 g
(We divide by 2 because we need 2 atoms of AgNO3 to mix with each BaCl2.