What is the percent yield if 4.65 grams of copper is produced when 1.897 grams of aluminum reacts with an excess of copper (ii) sulfate?
2al + 3cuso4 (aq) —> al2(so4)3 (aq) + 3cu

The percentage yield if 4.65 g of copper is produced will be 70.4 %.

Explanation:

The balanced equation is given as

                  2Al + 3CuSO4 > Al2(SO4)3 + 3Cu

1.87 g of Al =  0.0693 moles of Al *  3 moles of Cu / 2 moles of Al

                  = 0.10396 moles of Cu.

This corresponds to 6.606 g of Cu which refers to theoretical yield.

% yield = (actual / theoretical) * 100

            = (4.65 / 6.606) * 100

% yield = 70.4 %.  

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.

answermeridional ;