Respuesta :
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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(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:
[tex]\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol[/tex]
- To calculate the moles of a solute, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
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:
[tex]0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol[/tex]
- For the given chemical equation:
[tex]2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)[/tex]
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 = [tex]\frac{1}{2}\times 0.01925=0.009625moles[/tex] 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 = [tex]\frac{1}{2}\times 0.01925=0.009625moles[/tex] 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:
[tex]0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g[/tex]
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.
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
[tex] \large {\boxed {\boxed {\bold {mol = \frac {mass} {molar \: mass}}}} [/tex]
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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