A strip of zinc medal weighing 2.00 grams is placed in an aqueous solution containing 150.ml of .750M Tin(II) nitrate. Determine the grams of tin metal that will be precipitated.
The reaction is as followed:
Zn + Sn(NO3)2 --> Sn + Zn(NO3)2
Convert 0.750M Sn(NO3)2 to moles:
Molarity (M)= moles/volume (L)
Molarity*Volume (L)=moles
moles of Sn(NO3)2=0.750M*0.150L
This is similar to the limiting reagent problem. Proceed as you would do for the limiting reagent problem.
To determine the grams of tin metal that will be precipitated, we need to use stoichiometry and the balanced chemical equation of the reaction between zinc and tin(II) nitrate.
The balanced chemical equation is:
Zn + Sn(NO3)2 → Zn(NO3)2 + Sn
The stoichiometric ratio between zinc and tin in this reaction is 1:1, meaning that 1 mole of zinc will react with 1 mole of tin.
First, let's calculate the number of moles of tin(II) nitrate present in the solution:
Number of moles = concentration (in M) × volume (in L)
Given that the solution has a concentration of 0.750 M and a volume of 150 mL (or 0.150 L):
Number of moles of Sn(NO3)2 = 0.750 M × 0.150 L = 0.1125 moles
Since the stoichiometric ratio between zinc and tin is 1:1, we know that 0.1125 moles of tin(II) nitrate will react with an equal amount of zinc. So, the number of moles of tin that will be precipitated is also 0.1125 moles.
Finally, let's calculate the grams of tin metal that will be precipitated:
Grams = moles × molar mass
The molar mass of tin (Sn) is 118.71 g/mol.
Grams of tin = 0.1125 moles × 118.71 g/mol = 13.38 grams
Therefore, approximately 13.38 grams of tin metal will be precipitated.