What is the volume of 0.0450mol/L Ag^+(aq) necessary to react exactly 18.5 of Zn(s)?
To determine the volume of 0.0450 mol/L Ag+(aq) necessary to react exactly with 18.5g of Zn(s), we can use stoichiometry and the balanced chemical equation for the reaction between Ag+(aq) and Zn(s).
The balanced chemical equation is: 2Ag+(aq) + Zn(s) -> Zn2+(aq) + 2Ag(s)
From the balanced equation, we can see that 2 moles of Ag+(aq) are required for every 1 mole of Zn(s) that reacts. Therefore, the molar ratio between Ag+(aq) and Zn(s) is 2:1.
First, we need to determine the number of moles of Zn(s) in 18.5g.
Number of moles of Zn(s) = mass / molar mass
Molar mass of Zn(s) = 65.38 g/mol (from periodic table)
Number of moles of Zn(s) = 18.5g / 65.38 g/mol = 0.2823 mol
Since the molar ratio between Ag+(aq) and Zn(s) is 2:1, we need to have half the amount of Ag+(aq) compared to Zn(s) for the reaction to be complete.
Number of moles of Ag+(aq) = 0.2823 mol / 2 = 0.1411 mol
Finally, we can calculate the volume of the 0.0450 mol/L Ag+(aq) solution needed using its molarity:
Volume of Ag+(aq) = number of moles / molarity = 0.1411 mol / 0.0450 mol/L = 3.14 L
Therefore, approximately 3.14 liters of the 0.0450 mol/L Ag+(aq) solution is necessary to react exactly with 18.5g of Zn(s).