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

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

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Answer 1 · 2016-01-11T13:58:02

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

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?

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

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Explanation:

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If 9.01 g of BeBe reacts with 70.90 g of Cl_2Cl_2, how much BeCl_2BeCl_2 will be produced?

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Explanation:

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