Question

PLEASE ANSWER 30 POINTS!!!!!

How many grams of Ag would be produced from the complete reaction of 48 grams of Mg?
Mg + 2AgNO3 ----> 2Ag + Mg(NO3)2
Mg: 24 g/mol Ag: 108 g/mol
48g Mg --> g Ag

Answer 1

432 grams of Ag

Step-by-step explanation:

First, we need to determine the limiting reagent between Mg and AgNO3.Using the stoichiometry of the balanced chemical equation, we can see that 1 mole of Mg reacts with 2 moles of AgNO3 to produce 2 moles of Ag.

The number of moles of Mg present in 48 grams can be calculated as:

48 g / 24 g/mol = 2 moles Mg

Now, let's calculate the number of moles of Ag that can be produced from 2 moles of Mg:

2 moles Mg x (2 moles Ag / 1 mole Mg) = 4 moles Ag

Finally, we can calculate the mass of Ag produced by multiplying the number of moles of Ag by its molar mass:

4 moles Ag x 108 g/mol = 432 grams Ag

Therefore, 48 grams of Mg will produce 432 grams of Ag in this reaction.