Chemistry, 09.07.2019 16:20 heroicblad

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?

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Answer from: LilLappyLOL

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,

$Number \ of \ moles = \frac{Mass}{Molar \ mass}$

For AgNO₃

Mass = 4.22 g

Molar mass = 169.87 g/mol

∴ $Number \ of \ moles = \frac{4.22}{169.87}$

Number of moles of AgNO₃ present = 0.02484 moles

For AlCl₃

Mass = 7.73 g

Molar mass = 133.34 g/mol

∴ $Number \ of \ moles = \frac{7.73}{133.34 }$

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 $\frac{0.02484}{3}$ moles of AlCl₃ completely.

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

$\frac{0.02484}{3} = 0.00828$

∴ 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

$Mass = Number \ of \ moles \times Molar \ mass$

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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Answer from: Skatetank

3.6 grams

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of silver nitrate}=\frac{4.22g}{170g/mol}=0.025moles$

$\text{Number of moles of aluminium chloride}=\frac{7.73g}{133g/mol}=0.06moles$

$3AgNO_3(aq)+AlCl_3(aq)\rightarrow Al(NO)_3)_3(aq)+3AgCl(s)$

According to stoichiometry:

3 mole of AgNO_3 react with 1 mole of AlCl_3

0.025 moles of AgNO_3 react with = $\frac{1}{3}\times 0.025=8.3\times 10^{-3}moles$ of AlCl_3

Thus AgNO_3 is the limiting reagent as it limits the formation of product and AlCl_3 is the excess reagent.

3 moles of AgNO_3 produce = 3 moles of AgCl

0.025 moles of AgNO_3 produce = $\frac{3}{3}\times 0.025=0.025moles$ of AgCl

Mass of AgCl produced= $moles\times {\text {Molar mass}}=0.025\times 143=3.6g$

Thus 3.6 g of AgCl is produced.

Answer from: Quest

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Answer from: Quest

the answer is c.) 1575 j.

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Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl...