Question

What is the percent yield of lithium hydroxide from a reaction of 7.40 grams of lithium with 10.2 grams of water? The actual yield was measured to be 12.1 grams.

2Li(s) + 2H₂O(l) ⇒ 2LiOH(aq) + H₂(g)
Answers:
a 89.3%
b 47.4%
c 53.1%
d72.5%

Answer 1

The percent yield of lithium hydroxide from the reaction is 47.6%.

Step-by-step explanation:

To find the percent yield of lithium hydroxide, we first need to calculate the theoretical yield of lithium hydroxide.

From the balanced chemical equation, we see that 2 moles of lithium react with 2 moles of water to produce 2 moles of lithium hydroxide. Therefore, the molar ratio between lithium and lithium hydroxide is 1:1.

Using the molar mass of lithium (6.94 g/mol), we can convert the given mass of lithium (7.40 g) to moles: 7.40 g / 6.94 g/mol = 1.06 mol.

Since the molar ratio is 1:1, the theoretical yield of lithium hydroxide is also 1.06 mol.

To convert moles of lithium hydroxide to grams, we use the molar mass of lithium hydroxide (23.95 g/mol): 1.06 mol * 23.95 g/mol = 25.37 g.

Finally, we can calculate the percent yield by dividing the actual yield (12.1 g) by the theoretical yield (25.37 g) and multiplying by 100%: (12.1 g / 25.37 g) * 100% = 47.6%.