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geniusboy [140]
3 years ago
15

C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

Chemistry
1 answer:
melomori [17]3 years ago
8 0

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂  +  2HCl     →     PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number  of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

                            PbCl₂        :          Pb(NO₃)₂

                               1             :             1

                             0.23         :            0.23

Mass of Pb(NO₃)₂:

Mass = number of moles ×  molar mass

Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

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Can someone please help me with this question also explain the answers I am so confused thank you.
Archy [21]

The theoretical yield of H₂S is 13.5 g.

The percent yield is 75.5 %.

<h3>What is the theoretical yield of H₂S from the reaction?</h3>

The equation of the reaction is given below:

  • FeS + 2 HCl → FeCl₂+ H₂S

Moles of FeS reacting = mass/molar mass

Molar mass of FeS = 88 g/mol

Moles of FeS reacting = 35/88 = 0.398 moles

Moles of H₂S produced = 0.398 moles

Molar mass of H₂S = 34 g/mol

Mass of H₂S produced = 0.398 * 34 = 13.5 g

Theoretical yield of H₂S is 13.5 g.

  • Percent yield = actual yield/theoretical yield * 100%

Actual yield of H₂S = 10.2 g

Percent yield = 10.2/13.5 * 100%

Percent yield = 75.5 %

In conclusion, the actual yield is less than the theoretical yield.

Learn more about percent yield at: brainly.com/question/8638404

#SPJ1

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