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If 48.5 grams of aluminum metal (Al) react with 62.8 grams of sulfur (S) in a synthesis reaction, how many grams of the excess reactant will be left over when the reaction is complete?

Be sure to write out the balanced equation for this reaction and to show all of your work.

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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!

Answer:

13.122 grams of Al will be left.

Explanation:

The balanced possible reaction between Al and S is:

[tex]2Al+3S--->Al_{2}S_{3}[/tex]

Thus two moles of aluminium will react with three moles of sulfur

the moles of sulfur present = [tex]\frac{mass}{atomicmass}=\frac{62.8}{32}=1.96mol[/tex]

Moles of Al present = [tex]\frac{mass}{atomicmass}=\frac{48.5}{27}=1.796[/tex]

1.96 moles of sulfur will react with = [tex]\frac{2X1.96}{3}=1.31[/tex]moles of Al

The moles of Al left = The moles of excess reagent left = 1.796-1.31=0.486mol

the mass of Al left = moles X atomic mass =0.486X27=13.122 grams

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