A 0.652-g sample of a pure strontium halide reacts with

Chapter 3, Problem 98P

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid, and the solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. What is the formula of the original halide?

Questions & Answers

QUESTION:

A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid, and the solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. What is the formula of the original halide?

ANSWER:

Step 1 of 2

Here, we are going to determine the formula of the original halide present in the reaction.

The given equation of the reaction is

Given that,

Mass of  = 0.652 g

Mass of  = 0.755 g

Therefore,

The moles of   = (0.755 g )

= 0.0041 mol

Thus,

0.652 g  = 0.0041 mol

Thus, the molar mass of  =  = 158.630 g/mol

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back