1. How many grams of CO2 would be formed from 30 grams of C4H10 in the following (UNBALANCED) reaction? C4H10 + O2 → CO2 + H2O





2. How many moles of Ca3(PO4)2 would be formed from 40 grams of Ca in the following (UNBALANCED) equation?
Ca + H3PO4 → Ca3(PO4)2 + H2

Right no one here is just going to straight off answer your homework for you.

Can you balance equations? If so do that first if not post back and I'll help, IF you show what you have managed to do so far.

I can help but you need to show some initiative first.

1. (balanced equation) 2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

2. (balanced) 3 Ca + 2 H3PO4 = Ca3(PO4)2 + 3 H2

Thank you!

1. (balanced equation) 2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

2. (balanced) 3 Ca + 2 H3PO4 = Ca3(PO4)2 + 3 H2



Cool, cool, balancing equations is a really easy way to gain marks in exams! :D

I'm going to work through question 1 explaining what I am doing and then see if you can do question 2 on your own.
Basically what your balanced equation means is
2 moles of butane + 13 moles of oxygen  8 moles of carbon dioxide + 10 moles of water.
Therefore 2 moles of butane gives 8 moles of carbon dioxide
Or 1 mole butane gives 4 moles CO2
(you can ignore the others for now)
The molecular weights (grams/mole) are
Butane = 58, CO2 = 44
Grams = moles x molecular weight
So
Moles of butane = 30 (grams) / 58 (grams/mole)= 0.52 moles used
The ration from before was for every 1 mole butane you get 4 moles CO2
So 0.52 x 4 = 2.08 moles of CO2
You have been asked for the mass but you can figure that out easily as you have the equation, and you know the moles and molecular weight

Very helpful