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

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

Answer from: Quest

It shows that the experiment is designed properly.

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