workshop 4 f10 key - -3 moles XBr 2 XBr 2 mw is...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Workshop 4 1) Adapted from Exam 1, Fall 2007 (15 pts) A sample of a substance with the chemical formula XBr 2 (where X stands for an unknown cation of charge +2) weighs 0.5000 g. The substance is reacted with silver nitrate according to: XBr 2 + 2AgNO 3 2AgBr + X(NO 3 ) 2 The silver bromide produced is a solid and is found to weigh 1.0198 g. What is the atomic mass of X and which element is it? (Hint: you will need all the significant figures available) Have 5.430x10 -3 moles AgBr, so have 2.715 x10
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: -3 moles XBr 2 XBr 2 mw is 0.5000/2.715 x10-3 = 184.16g/mol X is Mg 2) Adapted from Exam 1, Fall 2005 What if I want to make 130 grams of iodine pentafluoride and I know the reaction proceeds with a 83% yield. How many grams of iodine and fluorine should I start with? 130g actual yield => 156g theoretical yield=> 0.706 moles iodine pentafluoride => 1.76 moles fluorine and 0.353 moles iodine => 66.9g fluorine and 89.6g iodine...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern