Only 3 questions, Have to post 1 after each answer,

(first)
http://media1.acellus.com/Library/00001127_630_A1.png

This post was edited by daiblorules on Feb 12 2014 05:25pm

Please.

so 1.46 moles of carbon + 2*1.46 moles of F2 will produce 1.46 moles of CF4

1.46 mole cf4 * molar mass of CF4 = 1.46 * 88.00 g = 128.48 g

Incorrect.

What are your needed sig figs?

with less rounding

88.0043 * 1.46 = 128.486278

This post was edited by cialda on Feb 12 2014 06:54pm

http://media1.acellus.com/Library/00001127_631_A1.png
Cialda was correct.

This post was edited by daiblorules on Feb 13 2014 12:00am

same thing here,
5.15 moles of CO + 3*5.15 mole H2 -> 5.15 H2O + 5.15 mole CH4

5.15 * molar mass of CH4 = 5.15 * 16.04 = 82.606 g

http://media1.acellus.com/Library/00001127_628_A1.png
Thank you bro

look at the formula first. its already balanced and you see that for 2 moles aluminum on reactant side you produce 2 moles aluminum chloride as a product.

this means if you produce 8.7 mole aluminum chloride you reacted 8.7 mole aluminum (with chlorine, which is irrelevant for this question, it is assumed you have unlimited chlorine since they don't mention a chlorine amount in problem)

so 8.7 moles aluminum x 1 mole aluminum is 26.982 g aluminum.

8.7x 26.982 = 234.7 grams aluminum needed