Calculate the volume of chlorine gas at STP that is required to completely react with 3.50 g of silver, using the following equation:
2Ag(s)+Cl2(g)->2AgCl(s)
First, we need to find the number of moles of silver (Ag) in 3.50 g.
Molar mass of Ag = 107.87 g/mol
Number of moles = Mass / Molar mass
Number of moles of Ag = 3.50 g / 107.87 g/mol = 0.0324 mol
From the balanced chemical equation, we see that 2 moles of Ag react with 1 mole of Cl2.
So, 0.0324 moles of Ag will react with 0.0162 moles of Cl2.
Now, we can use the ideal gas law to find the volume of Cl2.
PV = nRT
Given:
P = 1 atm (STP)
V = volume of Cl2
n = 0.0162 mol
R = 0.0821 L.atm/mol.K (gas constant for ideal gas law)
T = 273 K
V = nRT/P
V = (0.0162 mol)(0.0821 L.atm/mol.K)(273 K) / 1 atm = 0.360 L
Therefore, the volume of chlorine gas at STP that is required to completely react with 3.50 g of silver is 0.360 L.