2Al + 3CuSO4–>3Cu+ Al2(SO4)3 If 2.75 g Al reacted completely with excess CuSO4 and the percent yield of Cu is 80.5%, what mass of Cu was produced?

7.82 g of Cu

Explanation:

2 moles of Al react to 3 moles of copper sulfate in order to produce 3 moles of copper and 1 mol of aluminum sulfate.

Firstly we determine the moles of reactant.

As copper sulfate is in excess, Al is the limiting.

2.75 g . 1mol /26.98g = 0.102 moles

Ratio is 2:3. 2 moles of Al, can produce 3 moles of Cu

So the 0.102 moles of Al will produce(0.102 . 3) /2 = 0.153 moles.

We convert moles to mass: 0.153 mol . 63.5g /mol = 9.71 g

That's the theoretical yield (100 % yield reaction)

We know that: (yield produced / theoretical yield) . 100 = percent yield

We replace:

(Yield produced / 9.71g) . 100 = 80.5  %

(Yield produced / 9.71g) = 0.805

Yield produced = 0.805 .  9.71g = 7.82 g

6.4 cm (max) - 25 cm (min)

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.