If you want to produce 45.0 g of iron (III) nitrate, how much silver nitrate will you need to begin with?
3 AgNO₃ + FeCl₃ ---> 3 AgCl + Fe(NO₃)₃
According to the balanced equation, 3 moles of AgNO₃ when reacting with excess FeCl₃, will produce 1 mol of Fe(NO₃)₃. We can use that relationship to find the number of moles of Ag(NO₃)₃ necessary to produce 45.0 g of Fe(NO₃)₃. But first we have to find the number of moles of Fe(NO₃)₃ that we have in those 45.0 g of it using the molar mass.
atomic mass of Fe = 55.85 amu
atomic mass of N = 14.01 amu
atomic mass of O = 16.00 amu
molar mass of Fe(NO₃)₃ = 55.85 + 3 * 14.01 + 9 * 16.00
molar mass of Fe(NO₃)₃ = 241.88 g/mol
moles of Fe(NO₃)₃ = mass of Fe(NO₃)₃/molar mass of Fe(NO₃)₃
moles of Fe(NO₃)₃ = 45.0 g/(241.88 g/mol)
moles of Fe(NO₃)₃ = 0.186 moles
3 AgNO₃ + FeCl₃ ---> 3 AgCl + Fe(NO₃)₃
We said before that according to the coefficients of the reaction 3 moles of AgNO₃ will produce 1 mol of Fe(NO₃)₃. So we can use that relationship to find the number of moles of AgNO₃ that we need to produce 0.186 moles of Fe(NO₃)₃.
moles of AgNO₃ = 0.186 moles of Fe(NO₃)₃ * 3 mol of AgNO₃/(1 mol of Fe(NO₃)₃)
moles of AgNO₃ = 0.558 moles
And finally we can convert those moles into grams using the molar mass of AgNO₃.
atomic mass of Ag = 107.87 amu
atomic mass of N = 14.01 amu
atomic mass of O = 16.00 amu
molar mass of AgNO₃ = 1 *107.87 + 1* 14.01 + 3 * 16.00
molar mass of AgNO₃ = 169.88 g/mol
mass of AgNO₃ = moles of AgNO₃ * molar mass of AgNO₃
mass of AgNO₃ = 0.558 moles * 169.88 g/mol
mass of AgNO₃ = 94.8 g
Answer: if we want to produce 45.0 g of iron (III) nitrate we need 94.8 g of silver nitrate.
molar mass of Fe(NO₃)₃ = 241.88 g/mol
molar mass of AgNO₃ = 169.88 g/mol
3 AgNO₃ + FeCl₃ ---> 3 AgCl + Fe(NO₃)₃
45.0 g of Fe(NO₃)₃ * 1 mol of Fe(NO₃)₃/(241.88 g of Fe(NO₃)₃) * 3 moles of AgNO₃/(1 mol of Fe(NO₃)₃) * 169.88 g of AgNO₃/(1 mol of AgNO₃) = 94.8 g of
AgNO₃