Question

How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?

Balanced equation: AgNO3 + LiOH = AgOH + LiNO3

Answer 1

AgNO3+Li-->LiNO3+Ag

It is already balanced.

And turns out to be M AgNO3=577.6mg

Hope this helps!

Answer 2

`m_(AgNO_3)=577.6mg`

Step-by-step explanation:

Hello there!

In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:

`n_(LiOH)=81.5mg*(1g)/(1000mg)*(1mol)/(23.95g) =0.0034molLiOH`

Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:

`m_( AgNO_3)=0.0034molLiOH*(1molAgNO_3)/(1molLiOH) *(169.87gAgNO_3)/(1molAgNO_3) *(1000gAgNO_3)/(1gAgNO_3) \\\\m_(AgNO_3)=577.6mg`

Best regards!