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exis [7]
3 years ago
14

In this reaction, what roll does the lead (II) nitrate play when 50.0 mL of 0.100M iron (III) chloride are mixed with 50.0 mL of

0.100M lead (II) nitrate?
Chemistry
2 answers:
scoundrel [369]3 years ago
8 0
Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
     2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)   
Calculating the amount of the precipitate lead(II) chloride each reactant will produce: 
     mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
                       = 0.00500mol PbCl2
     mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3)                                 = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
Jobisdone [24]3 years ago
4 0

Answer: iron (III) chloride is the excess reactant in the reaction.

Explanation:

i just did the assignment

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3 0
3 years ago
Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride
IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1\text{ and }n_1 are the volume and number of moles of methane gas

V_2\text{ and }n_2 are the volume and number of moles of hydrogen chloride

We are given:

V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol

Putting values in above equation, we get:

\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

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Answer:

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Explanation:

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