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

Jan 11, 2016

#### Answer:

$B e \left(s\right) + C {l}_{2} \left(g\right) \rightarrow B e C {l}_{2} \left(s\right)$

#### 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 $B e C {l}_{2}$ would we isolate upon complete reaction. My guess is about 80 g. Can you refine this estimate?