What mass of silver can be produced by reacting 0.300 L of 0.100 M AgNO3 with an excess of Mg?
.1 M AgNO3 equals .1M Ag
convert to moles, then to grams
Mg + 2AgNO3 ==> 2Ag + Mg(NO3)2
mols AgNO3 = M x L
mols Ag = mols AgNO3
g Ag = mols Ag x atomic mass Ag.
To determine the mass of silver (Ag) produced in the given reaction, we need to use stoichiometry. The balanced chemical equation for the reaction between silver nitrate (AgNO3) and magnesium (Mg) is:
2AgNO3 + Mg -> Mg(NO3)2 + 2Ag
According to the equation, for every 2 moles of AgNO3, 2 moles of Ag are produced. Therefore, we can use the relationship between moles and concentration to find the number of moles of AgNO3 reacting:
moles of AgNO3 = volume of AgNO3 solution (in L) * concentration of AgNO3 (in M)
moles of AgNO3 = 0.300 L * 0.100 M = 0.030 moles of AgNO3
Since the stoichiometry tells us that 2 moles of Ag are produced for every 2 moles of AgNO3, we can conclude that 0.030 moles of AgNO3 will produce 0.030 moles of Ag.
To find the mass of Ag produced, we need to multiply the number of moles of Ag by its molar mass, which is found on the periodic table. The molar mass of Ag is approximately 107.87 g/mol.
mass of Ag = moles of Ag * molar mass of Ag
mass of Ag = 0.030 moles * 107.87 g/mol ≈ 3.23 g
Therefore, approximately 3.23 grams of silver can be produced by reacting 0.300 L of 0.100 M AgNO3 with an excess of Mg.