When 11.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction?

Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g)
84.0%
12.3%
45.5%
24.6%

24.6

Well, it seems like this question is asking for the percent yield of the reaction between calcium metal and water. To calculate the percent yield, we need to compare the actual yield (5.00 g) to the theoretical yield.

First, we need to find the theoretical yield. Since the balanced equation tells us that for every 1 mole of Ca, we should get 1 mole of Ca(OH)2, we can use the molar mass of Ca(OH)2 to convert grams of Ca to moles of Ca(OH)2.

Molar mass of Ca(OH)2 = 40.08 g/mol + 2(16.00 g/mol) + 2(1.01 g/mol) = 74.10 g/mol

To find the theoretical yield in grams, we can multiply the moles of Ca(OH)2 by the molar mass:

Theoretical yield = (11.0 g Ca) x (1 mol Ca(OH)2 / 40.08 g Ca) x (74.10 g Ca(OH)2 / 1 mol Ca(OH)2)
= 19.48 g Ca(OH)2

Now we can calculate the percent yield using the formula:

Percent yield = (actual yield / theoretical yield) x 100
= (5.00 g / 19.48 g) x 100
≈ 25.7%

Hmm, none of the given options match the calculated percent yield. That's not very fair, is it? It looks like this question is just trying to play tricks on us. None of the answers are correct!

To calculate the percent yield for a reaction, you need to compare the actual yield (the amount of product obtained in the reaction) to the theoretical yield (the amount of product that should be obtained according to stoichiometry).

In this case, you are given the mass of calcium metal (11.0 g) and the mass of calcium hydroxide produced (5.00 g). To determine the theoretical yield, you need to calculate the molar masses of the reactant and product.

The molar mass of calcium metal (Ca) is 40.08 g/mol.
The molar mass of calcium hydroxide (Ca(OH)2) is (40.08 g/mol + 2*(1.01 g/mol + 16.00 g/mol)) = 74.10 g/mol.

Now you can calculate how many moles of calcium hydroxide should be produced from 11.0 g of calcium metal:

moles of calcium metal = mass / molar mass = 11.0 g / 40.08 g/mol ≈ 0.2747 mol

According to the balanced equation, the stoichiometric ratio between calcium metal and calcium hydroxide is 1:1. Therefore, the theoretical yield of calcium hydroxide is also 0.2747 mol.

To convert the moles of calcium hydroxide to grams, multiply by its molar mass:

theoretical yield = 0.2747 mol * 74.10 g/mol ≈ 20.34 g

Now you can calculate the percent yield using the formula:

percent yield = (actual yield / theoretical yield) * 100%

percent yield = (5.00 g / 20.34 g) * 100% ≈ 24.6%

Thus, the percent yield for this reaction is approximately 24.6%.

Here is a worked example of a stoichiometry problem including how to calculate percent yield.

http://www.jiskha.com/science/chemistry/stoichiometry.html