For the reaction,  3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O, how many
grams of Cu would be needed to react with 4.0 mol of nitric acid?
a. 95.3g
b. 63.5g
c. 191g
d. 1.50g

Hi again!!

A little concept:
A balanced equation also shows a macroscopic quantitative
relationship.
Example:
C3H8 + 5 O2 ---> 3 CO2 + 4 H2O 
This balanced reaction equation shows that 5 moles of oxygen reacts
with 1 mole of propane generating 3 moles of carbon dioxide and 4
moles of water.

For the following reaction:
3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
We can conclude that we need 3 moles of Cu to react with 8 moles of
nitric acid (HNO3). So it is easy to see that if we have "only" 4
moles of HNO3 we will need "only" 1.5 moles of Cu to react with them.
Now we need to know how much grams are 1.5 moles of Cu. The atomic
weight of the Cu is 63.546, then the weight of one mol of Cu will be
63.546g; and 1.5 moles of Cu will weight 63.546g x 1.5 = 95.319g .

The correct answer is a.

Hope this helps.

Regards.
livioflores-ga

---

Clarification of Answer by livioflores-ga on 07 Jul 2003 22:58 PDT

This page will be useful:
"Reaction Equations" from the Faculty of Science Website of the
University of Waterloo, Ontario, Canada:
http://www.science.uwaterloo.ca/~cchieh/cact/c120/reaction.html

Good Luck!!

livioflores-ga