Answer:
lead ii nitrate is the answer
This is the balanced eq
N2 + 3H2 -> 2NH3
first you need to find mole of N2 by using
mol = mass ÷ molar mass.
mol N2= 20g ÷ (14.01×2)g/mol
=0.7138mol
then look at the coefficient between H2 and NH3.
it is N2:NH3
1:2
0.7138:0.7138×2
0.7138:1.4276 moles
moles of NH3 = 1.4276 moles
So you have evidence that the experiment is true or correct.
Balance each one by adding electrons to make the charges on both sides the same:
Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag
Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:
2 Ag+ + 2 e- --> 2 Ag
Now, just add the two half reactions together, cancelling anything that's the same on both sides:
2 Ag+ + Sn --> Sn2+ + 2 Ag
And you're done.
