What is the mass of copper produced if 4.87 g of #Al# reacts with copper sulfate in the reaction #2Al + 3CuSO_4 -> Al_2(SO_4)_3 + 3Cu#?

Answer 1

Approx. #17# #g# copper metal result.

#2Al + 3CuSO_4 rarr Al_2(SO_4)_3 + 3Cu#
#"Moles of aluminum " =(4.87*g)/(26.98*g*mol^-1)# #=# #0.181# #mol#
From the equation, we know that #3# moles of copper are reduced per #2# mol aluminum oxidized.
So #3/2xx0.181*mol xx 63.55*g*mol^-1# #=# #?g#
You have given the stoichiometry. It explicitly says that 3 moles of copper salt are reduced per 2 moles of aluminum oxidized. So simply multiply the molar quantity of aluminum by #3/2#.
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Answer 2

To find the mass of copper produced, first, calculate the moles of aluminum used. Then, use the mole ratio from the balanced equation to determine the moles of copper produced. Finally, use the molar mass of copper to find the mass of copper produced.

  1. Calculate moles of Al: Moles of Al = mass of Al / molar mass of Al

  2. Use the mole ratio from the balanced equation to find moles of Cu produced: Moles of Cu = (moles of Al) * (3 moles of Cu / 2 moles of Al)

  3. Calculate the mass of Cu produced: Mass of Cu = moles of Cu * molar mass of Cu

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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