Solving Lead(ll) silver nitrate

In mass to mass example

Stoichometry

How to solve for 12.27 gPb(NO3)2

First of all you start with your given in the top left for example 12.27 g Pb(NO3)2, onto the bottom of the 2nd square you put in the molar mass of Pb(NO3)2 =331.2 g/mol on top of that you put 1 mole Pb(NO3)2 because that 1 mole is the same mass of the lead nitrate. Then you move that 1 mole lead nitrate because of their coefficients and do the same on the top box for silver which is 2 moles (since the coefficient is that on the product side) then aside that top box is the molar mass of silver which is 107.868 g/mol the bottom is 2 mole Ag then you multiple the top and divide by the bottom. The answer is 7.99 mol of silver

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The top is another example of how the format of it looks like.