Question

If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3, what is the mass of the AgCl precipitate?

Answer 1

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.

What is Stoichiometry ?

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.

What is Balanced Chemical Equation ?

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.

= $\frac{0.00385}{2}$

= 0.001925 mol                      

According to Stoichiometry Mass of AgCl

$= 0.0385 \times \frac{0.1}{1\ \text{mol}} \times \frac{2\ \text{mol} AgCl}{2\ \text{mol} AgNO_3} \times \frac{143.4\ g}{1\ \text{mol}}$

= 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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