Question

A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Write the balanced chemical equation for the reaction. Physical states are optional. chemical equation: What is the limiting reactant? calcium hydroxide hydrochloric acid How many grams of salt are formed after the reaction is complete? mass of salt: gg How many grams of the excess reactant remain after the reaction is complete? excess reactant remaining:

Answer 1

Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.  

Explanation:

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

For calcium hydroxide:

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol$

• To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

$0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol$

• For the given chemical equation:

$2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)$

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

• Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

$0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g$

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

Answer 2

HCl is a limiting reactant

mass of salt: 1.068375 g

8.559 g  of the excess reactant (Ca(OH)₂)remain after the reaction is complete

Further explanation

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

Terms used:

Mole

The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

$\large {\boxed {\boxed {\bold {mol = \frac {mass} {molar \: mass}}}}$

We determine the mole of each reactant to determine the limiting reactant

then:

mol Ca (OH₂ = mass: molar mass

mole of Ca (OH)₂ = 9.27 g: 74

mole of Ca (OH)₂ = 0.1253

mole HCl: 38.5 ml x 0.5 M = 19.25 mlmol = 0.01925 mol

From the number of moles, it can be seen that HCl is a limiting reactant

Reaction:

                     Ca(OH)₂  +    2HCl         ⇒    CaCl₂      +         2H₂O

initial mole    0.1253         0.01925

reaction        0.009625   0.01925        0.009625           0.01925

remaining     0.115675       -                  0.009625           0.01925

Remaining unreacted Ca (OH)₂ mole: 0.115675

Amount of mass of CaCl₂ salt formed:

CaCl₂ mass = mole x molar mass

CaCl₂ mass = 0.009625 x 111

CaCl₂ mass = 1.068375

Remaining Ca(OH)₂ = 0.115675 x 74 = 8.559 g

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