Suppose that magnesium would react exactly the same as copper in this experiment. How many grams of magnesium would have been used in the reaction if 1.000 g of silver were produced? The atomic mass of magnesium is 24.31 g/mol, and the atomic mass of silver is 107.87 g/mol.
What experiment????
To solve this problem, we need to use stoichiometry and the concept of molar ratios.
First, let's write a balanced chemical equation for the reaction:
2Ag + Mg -> Mg2+ + 2Ag+
According to the balanced equation, for every 2 moles of Ag produced, 1 mole of Mg is consumed.
Now, let's set up a conversion factor using the molar ratios:
1 mole Mg / 2 moles Ag = x moles Mg / 1.000 g Ag
To find the moles of Mg, we need to determine the moles of Ag first. We can do this by using the molar mass of Ag:
1.000 g Ag * (1 mole Ag / 107.87 g Ag) = 0.0092697 moles Ag
Now, let's plug this value into our conversion factor to find the moles of Mg:
(1 mole Mg / 2 moles Ag) = x moles Mg / 0.0092697 moles Ag
x = (1 mole Mg / 2 moles Ag) * 0.0092697 moles Ag
x = 0.00463485 moles Mg
Finally, we can calculate the grams of Mg using the molar mass of Mg:
0.00463485 moles Mg * (24.31 g Mg / 1 mole Mg) = 0.1128 g Mg
Therefore, approximately 0.1128 grams of magnesium would have been used in the reaction if 1.000 g of silver were produced.