Answer: Your balanced equation is 2Al + 3S --> Al_2S_3. In two separate equations, first use the 48.5g of Al to find the amount (in moles) of Al_2S_3. Then use the 62.8g of S to find the amount (in moles) of Al_2S_3. By doing these, you find which reactant (the Al or the S) produces the least amount of Al_2S_3. Since it's the S, that's the limiting reactant: you then use that 62.8g of S to find how many grams of Al will be used in the reaction. (Use the same balanced equation.) You find that 35.2g of Al will be used, then find the excess by subtracting that 35.19 from the original 48.5g. Your excess amount is 13.3g of Al!
Explanation:
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The amount of excess reactant that left over the chemical reaction between aluminum and ammonium perchlorate is 0.9 moles.
Mass into moles (n) will be converted by using the below equation as:
n = W/M, where
W = given mass
M = molar mass
Moles of 45g of Aluminum = 45g / 27g/mol = 1.66 mol
Moles of 301g of ammonium perchlorate = 301g / 117.4g/mol = 2.56 mol
Both reactants are reacted with each other and one of the reactant is fully consumed, so the excess reactant will be calculated as:
Excess reactant = 2.56mol - 1.66mol = 0.9 moles
Hence amount of excess reactant is 0.9 moles.
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