Question

When 10.0 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)

Answer 1

Answer:

27.03 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For calcium

Given mass = 10.0 g

Molar mass of calcium = 40.078 g/mol

Moles of calcium = 10.0 g / 40.078 g/mol = 0.2495 moles

According to the given reaction:

$Ca_{(s)}+2H_2O_{(l)}\rightarrow Ca(OH)_2_{(aq)}+H_2_{(g)}$

1 mole of calcium on reaction forms 1 mole of calcium hydroxide

Thus,

0.2495 moles of calcium on reaction forms 0.2495 mole of calcium hydroxide

Moles of calcium hydroxide = 0.2495 moles

Molar mass of calcium hydroxide = 74.093 g/mol

Thus, Mass = Moles * Molar mass = 0.2495 moles * 74.093 g/mol = 18.5 g

Theoretical yield = 18.5 g

Given experimental yield = 5.00 g

% yield = (Experimental yield / Theoretical yield) × 100 = (5.00/18.5) × 100 = 27.03 %