Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?

The mass in grams of AgCl produced when 4.22 g of AgNO₃ react with 7.73 g of AlCl₃ is 3.56 g

From the question, the equation of the reaction is given as

3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s)

This can be written properly as

3AgNO₃ (aq) + AlCl₃ (aq) → Al(NO₃)₃ (aq) + 3AgCl (s)

From the balanced chemical equation above, we observe that

3 moles of AgNO₃ will react with 1 mole of AlCl₃ to yield 1 mole of Al(NO₃)₃ and 3 moles of AgCl.

To determine the mass of AgCl that would be produced when 4.22 g of AgNO₃ react with 7.73 g of AlCl₃,

First, we will convert the given masses to number of moles

Using the formula,

For AgNO₃

Mass = 4.22 g

Molar mass = 169.87 g/mol

∴

Number of moles of AgNO₃ present = 0.02484 moles

For AlCl₃

Mass = 7.73 g

Molar mass = 133.34 g/mol

∴

Number of moles of AlCl₃ present = 0.05797 moles

Since,  

3 moles of AgNO₃ react with 1 mole of AlCl₃

Then,

0.02484 moles of AgNO₃ will react with moles of AlCl₃ completely.

(NOTE: AgNO₃ is the limiting reagent and AlCl₃ is the excess reagent)

∴ Only 0.00828 moles of AlCl₃ will react.

Now, since

3 moles of AgNO₃ will react with 1 mole of AlCl₃ to yield 1 mole of Al(NO₃)₃ and 3 moles of 3AgCl

Therefore,

0.02484 moles of AgNO₃ will react with 0.00828 mole of AlCl₃ to yield 0.00828 mole of Al(NO₃)₃ and 0.02484 moles of AgCl

∴ 0.02484 moles of AgCl is produced during the reaction.

Now, we will convert this amount to grams

From the formula

Number of moles = 0.02484 moles

Molar mass = 143.32 g/mol

∴ Mass of AgCl produced = 0.02484 × 143.32

Mass of AgCl produced = 3.56 g

Hence, the mass in grams of AgCl produced when 4.22 g of AgNO₃ react with 7.73 g of AlCl₃ is 3.56 g

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3.6 grams

Explanation:

According to stoichiometry:

3 mole of react with 1 mole of

0.025 moles of react with = of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

3 moles of produce = 3 moles of

0.025 moles of produce = of

Mass of produced=

Thus 3.6 g of is produced.