When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2 Al + 6HCl → 2 AlCl3 + 3 H2

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When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2 Al + 6HCl → 2 AlCl3 + 3 H2

Chemistry
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I know I need to balance the equation first.
6Al + 18HCl -> 6AlCl3 + 9H2
It was already balanced to begin with
ok so how do i find limiting reactant and # of moles of H2 that can be formed
You find the limiting reagent by finding which ever one needs the least amount of moles to produce your product. Keeping in mind that your coefficients need to be taken into account
|dw:1435565108958:dw|
so HCL is limiting the reaction as it need to be more to react with all the AL , the current amount can only react with approximately 4.3 moles. Now we have to solve for H2 which will be formed.
To do so we will make use of HCl ( the least amount) or we will get excessive number more than we will really get. |dw:1435565479220:dw|
so we have 6 - 4.3 = 1.7 moles excessive of AL and 6.5 moles of H2 will be formed

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