Here's the question:
"What mass of the excess reactant will remain when the reaction is complete?
2Al(s) + 3Cl2(g) = Al2Cl6(s)
Given:
0.100 mol Al
0.0571 mol Cl2
Cl2 is the limiting reagent
5.08g Al2Cl6 produced"
What do I do to figure out with this information to find the mass in grams of the excess reactant? What is the formula I would use?
Any chem experts wanna share their knowledge with me?
"What mass of the excess reactant will remain when the reaction is complete?
2Al(s) + 3Cl2(g) = Al2Cl6(s)
Given:
0.100 mol Al
0.0571 mol Cl2
Cl2 is the limiting reagent
5.08g Al2Cl6 produced"
What do I do to figure out with this information to find the mass in grams of the excess reactant? What is the formula I would use?
Any chem experts wanna share their knowledge with me?
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It is simpler knowing Cl2 is limiting.
Just see how many mol of Al would react with the given amount of Cl2
It is .0571 x [2/3] = you calculate it
Mols Al left over = .1 minus Al consumed
Then multiply by atomic mass Al to get mass
As a check, convert 5.08g Al2Cl6 into moles and double that to find how many mol Al were used, then subtract from .1
Just see how many mol of Al would react with the given amount of Cl2
It is .0571 x [2/3] = you calculate it
Mols Al left over = .1 minus Al consumed
Then multiply by atomic mass Al to get mass
As a check, convert 5.08g Al2Cl6 into moles and double that to find how many mol Al were used, then subtract from .1