Respuesta :
Answer: The amount of solid product (lead dichromate) formed is 0.025 moles
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
- For potassium dichromate:
Molarity of potassium dichromate solution = 0.25 M
Volume of solution = 100.0 mL
Putting values in equation 1, we get:
[tex]0.25M=\frac{\text{Moles of potassium dichromate}\times 1000}{100.0}\\\\\text{Moles of potassium dichromate}=\frac{0.25\times 100}{1000}=0.025mol[/tex]
- For lead nitrate:
Molarity of lead nitrate solution = 0.25 M
Volume of solution = 100.0 mL
Putting values in equation 1, we get:
[tex]0.25M=\frac{\text{Moles of lead nitrate}\times 1000}{100.0}\\\\\text{Moles of lead nitrate}=\frac{0.25\times 100}{1000}=0.025mol[/tex]
The chemical equation for the reaction of potassium dichromate and lead nitrate follows:
[tex]K_2Cr_2O_7(aq.)+Pb(NO_3)_2(aq.)\rightarrow PbCr_2O_7(s)+2KNO_3(aq.)[/tex]
The solid precipitate formed here is lead dichromate
By Stoichiometry of the reaction:
1 mole of lead nitrate produces 1 mole of lead dichromate
So, 0.025 moles of lead nitrate will produce = [tex]\frac{1}{1}\times 0.025=0.025mol[/tex] of lead dichromate
Hence, the amount of solid product (lead dichromate) formed is 0.025 moles
