If 9.01 g of #Be# reacts with 70.90 g of #Cl_2#, how much #BeCl_2# will be produced?

1 Answer
Jan 11, 2016

Answer:

#Be(s) + Cl_2(g) rarr BeCl_2(s)#

Explanation:

You have been given above a stoichiometric equation. #1# mole of beryllium metal is oxidized by #1# mole of chlorine gas to give 1 mole of beryllium chloride.

Given that we start with #1# mol of beryllium and #1# mole of chlorine gas, how much #BeCl_2# would we isolate upon complete reaction. My guess is about 80 g. Can you refine this estimate?