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stoichiometry of redox reactions

How many grams of magnesium are needed to displace all the silver from 25.0 g of silver nitrate?

I think that your supposed to write the redox equation first but I need help writing it out. Also do you have to use the 25.0 g of silver in order to get the grams of magnesium? Thanks

These are the half reactions. From these you can get the balanced equation.

$Mg \rightarrow Mg^{2+} + 2e^-$

$Ag^+ + e^- \rightarrow Ag$

Yes, you do need to use the 25.0 g of Ag in order to figure out the grams of magnesium. 1. From the mass of Ag, you determine the number of moles of Ag. w. Knowing the number of moles of Ag and the stoichiometry, you figure out the number of moles of Mg that are required. 3. From the number of moles of Mg you figure out the mass of Mg.

Quote:

Originally Posted by Perito

These are the half reactions. From these you can get the balanced equation.

$Mg \rightarrow Mg^{2+} + 2e^-$

$Ag^+ + e^- \rightarrow Ag$

Yes, you do need to use the 25.0 g of Ag in order to figure out the grams of magnesium. 1. From the mass of Ag, you determine the number of moles of Ag. w. Knowing the number of moles of Ag and the stoichiometry, you figure out the number of moles of Mg that are required. 3. From the number of moles of Mg you figure out the mass of Mg.

Is this right?

Mg --> Mg^2+ + 2e-
2(Ag+ + e- --> Ag)
---------------------------
Mg + 2Ag+ + 2e- ---> Mg2+ + 2e- + 2Ag
Mg + 2Ag+ --> Mg2+ + 2Ag

25.0g x
------ = --------------
107.9g x 2mol 24.3 g x 1 mol

x = 2.82g of Mg ( I'm not sure if i made a mistake )

Oh and I don't get why you didn't use AgNO3 in the half reactions.

Quote:

Originally Posted by lisanoce

Is this right?

Mg --> Mg^2+ + 2e-
2(Ag+ + e- --> Ag)
---------------------------
Mg + 2Ag+ + 2e- ---> Mg2+ + 2e- + 2Ag
Mg + 2Ag+ --> Mg2+ + 2Ag

25.0g x
------ = --------------
107.9g x 2mol 24.3 g x 1 mol

x = 2.82g of Mg ( im not sure if i made a mistake )

Oh and I don't get why you didnt use AgNO3 in the half reactions.

25.0g x
------ = --------------
107.9gx2mol 24.3g x 1 mol

x = 2.82g of Mg

Quote:

Originally Posted by lisanoce

25.0g x
------ = --------------
107.9gx2mol 24.3g x 1 mol

x = 2.82g of Mg

Ugh this is my first time using this sorry it keeps coming out wrong : S

25.0g/107.9g(2 mol) = x/24.3g(1 mol)

x= 2.82 g of Mg

Your equation is correct

$Mg + 2Ag^+ \rightarrow Mg^{2+} + 2Ag$

The rest of it is almost correct. You have 25 g of Ag(NO3). I probably misled you by mentioning only silver instead of silver nitrate. The problem specified 25 g of silver nitrate, so you need to use the molecular weight of silver nitrate, 169.87 g/mole.

$\frac {25\,g}{169.87 g/mole} = 0.147\, moles\, of\, Ag(NO_3)$
1
According to the equation, you need 1 mole of Mg for every 2 moles of Ag+ so

$moles\, of \, Mg\, required]\, =\, \frac {0.147}{2} = 0.736$
169.87

The atomic weight of Mg is 24.3050 so you need

$0.736\, moles \cdot \, 24.3050 \, \frac {g}{mole} = 1.789\, grams\, Mg$

The reason I didn't use Ag(NO3) in the half reactions, but only used Ag+ is that the NO3- group, in this case, doesn't do anything. It's merely a "spectator ion". You could have written

$2Ag(NO_3)+ Mg \rightarrow Ag + Mg(NO_3)_2$

and this would also have been correct.

in one equation, like you had, the math becomes

$\frac {25\, \cdot \, 24.305}{169.87 \, \cdot \, 2} = 1.7885$

unless I made a mistake.

Quote:

Originally Posted by lisanoce

How many grams of magnesium are needed to displace all the silver from 25.0 g of silver nitrate?

I think that your supposed to write the redox equation first but i need help writing it out. Also do you have to use the 25.0 g of silver in order to get the grams of magnesium? thanks

Ohh I get it now thanks!

Hang in there. It's fun once you learn it.