Question

What is the volume of 0.35 M aluminum bromide solution reacts with 150ml of 0.25M silver nitrate?
What is the mass of the products

Answer 1

First, you need to figure out and balance the equation
\[AlBr_{3}+3Ag(NO_{3}) -->Al(NO_{3})_{3}+3AgBr\]
Now you need to find the mass of aluminum nitrate and silver bromide using the givens in the reaction. Find which one is the limiting factor, then calculate based on that.

Answer 2

I dont know how to find the mass

Answer 3

Lets start with the first question about volume first. There are 22.4 L in a mole, so
\[\frac{ 150 mLAg(NO_{3})*L*mol*L*3molAg(NO_{3})*.35molAlBr_{3}*22.4L*1000mL}{ 1000mL*22.4L*.25molAg(NO_{3})*1molAlBr_{3}*L*mol*L }\]
I am using something called dimensional analysis. Just multiply the top row to get the first product, and divide it all by the product of the bottom row.
Hint: you can just multiply and divide the numbers. Tell me what you get.

Answer 4

I got L

Answer 5

I got 210 L