Solving Lead(ll) silver nitrate

In mass to mass example


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.