If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3. The mass of the AgCl precipitate is 0.552 g.
Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now we have to write the balanced equation
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
Molecular Weight of CaCl₂ = 110.98 g/mol
Molecular Weight of AgNO₃ = 170.01 g/mol
Molecular Weight of AgCl = 143.45 g/mol
Here,
Volume of CaCl₂ = 30.0 mL = 0.03L
Volume of AgNO₃ = 38.5 mL = 0.0385 L
Now find the number of moles
Number of moles = Volume × Molarity
Number of moles of CaCl₂ = 0.03 L × 0.150
= 0.00456 mol
Number of moles of AgNO₃ = 0.0385 L × 0.100
= 0.00385 mol
The stoichiometric ratio of AgNO₃ to CaCl₂ is 2:1.
= [tex]\frac{0.00385}{2}[/tex]
= 0.001925 mol
According to Stoichiometry Mass of AgCl
[tex]= 0.0385 \times \frac{0.1}{1\ \text{mol}} \times \frac{2\ \text{mol} AgCl}{2\ \text{mol} AgNO_3} \times \frac{143.4\ g}{1\ \text{mol}}[/tex]
= 0.552 g AgCl
Thus from the above conclusion we can say that If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3. The mass of the AgCl precipitate is 0.552 g.
Learn more about the Stoichiometry here: brainly.com/question/16060223
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