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What is the percent yield of calcium hydroxide when you have 83 grams of calcium chloride and 75 grams of water if the reaction prodcues 43.5 grams of calcium hydroxide. CaCl2+2H2O yields Ca(OH)2+2HCl

User GWR
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Final answer:

The percent yield of calcium hydroxide in this reaction is 79.7%.

Step-by-step explanation:

To calculate the percent yield of calcium hydroxide, we need to use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) × 100

The theoretical yield is the amount of calcium hydroxide that would be produced if the reaction went to completion. In this case, the balanced equation tells us that 1 mole of calcium chloride produces 1 mole of calcium hydroxide, so the theoretical yield of calcium hydroxide can be determined by converting the mass of calcium chloride to moles and then using the molar ratio between calcium chloride and calcium hydroxide.

Using the molar masses of calcium chloride (111 g/mol) and calcium hydroxide (74.1 g/mol), we can calculate the moles of calcium chloride:

Moles of CaCl₂ = Mass of CaCl₂ / Molar mass of CaCl₂

Moles of CaCl₂ = 83 g / 111 g/mol

= 0.747 moles

Since the reaction produces 1 mole of calcium hydroxide for every mole of calcium chloride, the theoretical yield of calcium hydroxide is also 0.747 moles.

Finally, we can calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100:

Percent Yield = (43.5 g / (0.747 mol × 74.1 g/mol)) × 100 = 79.7%

User Majid Shamkhani
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