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qaws [65]
3 years ago
6

An aqueous solution containing 7.96 g7.96 g of lead(II) nitrate is added to an aqueous solution containing 6.82 g6.82 g of potas

sium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: What is the limiting reactant? potassium chloride lead(II) nitrate The percent yield for the reaction is 89.3%89.3% . How many grams of precipitate is recovered? precipitate recovered: gg How many grams of the excess reactant remain? excess reactant remaining: g
Chemistry
1 answer:
Eduardwww [97]3 years ago
4 0

<u>Answer:</u> Lead nitrate is the limiting reagent and potassium chloride is the excess reagent. the amount of excess reagent left is 3.205 grams. The amount of precipitate (lead chloride) recovered is 5.96 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}   ....(1)

  • <u>For lead nitrate:</u>

Given mass of  lead nitrate = 7.96 g

Molar mass of  lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

\text{Moles of  lead nitrate}=\frac{7.96g}{331.2g/mol}=0.024mol

  • <u>For potassium chloride:</u>

Given mass of potassium chloride = 6.82 g

Molar mass of potassium chloride = 74.55 g/mol

Putting values in above equation, we get:

\text{Moles of potassium chloride}=\frac{6.82g}{74.55g/mol}=0.091mol

  • For the given chemical equation:

Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)

By Stoichiometry of the reaction:

1 mole of lead nitrate reacts with 2 moles of potassium chloride.

So, 0.024 moles of lead nitrate will react with = \frac{1}{2}\times 0.024=0.048moles of potassium chloride

As, given amount of potassium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, lead nitrate is considered as a limiting reagent because it limits the formation of product.

  • Amount of excess reagent (potassium chloride) left = 0.091 - 0.048 = 0.043 moles

Now, calculating the mass of potassium chloride from equation 1, we get:

Given mass of potassium chloride = 6.82 g

Molar mass of potassium chloride = 74.55 g/mol

Putting values in above equation, we get:

0.043mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=3.205g

By Stoichiometry of the reaction:

1 mole of lead nitrate produces 1 mole of lead chloride.

So, 0.024 moles of lead nitrate will produce = \frac{1}{1}\times 0.024=0.024moles of lead chloride

  • Now, calculating the mass of lead chloride from equation 1, we get:

Molar mass of lead chloride = 278.1 g/mol

Moles of lead chloride = 0.024 moles

Putting values in equation 1, we get:

0.024mol=\frac{\text{Mass of lead chloride}}{278.1g/mol}\\\\\text{Mass of lead chloride}=6.6744g

We are given:

Percentage yield of the reaction = 89.3 %

So, amount of lead chloride recovered will be = \frac{89.3}{100}\times 6.6744g=5.96g

Hence, lead nitrate is the limiting reagent and potassium chloride is the excess reagent. the amount of excess reagent left is 3.205 grams. The amount of precipitate (lead chloride) recovered is 5.96 g.

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The correct option is;

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Here, we have the reaction as follows;

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Therefore, one mole of Sn reacts with 2 moles HF to form one mole of SnF₂ and one mole of H₂

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Therefore, the number of moles of H₂ in 48 grams of H₂ is given by the relation;

Number \ of  \ moles \ of \, H_2 = \frac{Mass \ of  H_2}{Molar \ Mass \ of  H_2}=\frac{48}{2.01588} = 23.8109 \ moles \approx 24 \ moles

Since one mole each of SnF₂ and H₂ are produced, the number of moles of SnF₂ produced = 24 moles.

The number of moles of SnF₂ that will be produced is 24 moles.

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A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the ori
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4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

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Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

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To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

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