what js the percent yield of lithium hydroxide from a reaction of 7.40 g of lithium with 10.2 g of water? the actual yield was measured to be12.1 g

89%.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Li + 2H2O —> 2LiOH + H2

Next, we shall determine the masses of Li and H2O that reacted and the mass of LiOH produced from the balanced equation.

This is illustrated below:

Molar mass of Li = 7 g/mol

Mass of Li from the balanced equation = 2 x 7 = 14 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36 g

Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol

Mass of LiOH from the balanced equation = 2 x 24 = 48 g

Summary:

From the balanced equation above,

14 g of Li reacted with 36 g of H2O to produce 48 g of LiOH.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

14 g of Li reacted with 36 g of H2O.

Therefore, 7.4 g of Li will react with = (7.4 x 36)/14 = 19.03 g of H2O.

From the calculation made above, we can see that it will take a higher amount i.e 19.03 g than what was given i.e 10.2 g of H2O to react completely with 7.4 g of Li.

Therefore, H2O is the limiting reactant and Li is the excess reactant.

Next, we shall determine the theoretical yield of LiOH.

In this case we shall use the limiting reactant.

The limiting reactant is H2O and the theoretical yield of LiOH can be obtained as follow:

From the balanced equation above,

36 g of H2O reacted to produce 48 g of LiOH.

Therefore, 10.2 g of H2O will react to produce = (10.2 x 48)/36 = 13.6 g of LiOH.

Therefore, the theoretical yield of LiOH is 13.6 g

Finally, we shall determine the percentage yield of LiOH. This can be obtained as follow:

Actual yield = 12.1 g

Theoretical yield = 13.6 g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 12.1/ 13.6 x 100

Percentage yield = 89%

Therefore, the percentage yield LiOH is 89%.

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