If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
1 Answer
Pb(NO3)2- Lead Nitrate
NaI - Sodium Iodide
Moles of Lead(II) Nitrate =
= 25.0 grams//(207.20 xx 1 + 14.01 xx 2 + 16 xx 6
= 25.0 grams//331.22
= 0.075 moles
Moles of Sodium Iodide
= 15//(22.99+126.9)
= 15//149.89
= 0.100 Moles
Limiting Reagent in this case is Lead(II) Nitrate
Ratio of Moles of Lead Nitrate to Sodium Nitrate
1:2
Ratio of "moles" of lead nitrate to Sodium Nitrate
0.075 : x
Thus that means 0.150 moles of Sodium Nitrate
To calculate mass of Sodium Nitrate
Since we know that Moles =
Thus Mass =
Molar Mass of
Na - 22.99
N - 14.01
O - 3(16) = 48
85g/mol
Thus, Mass =
Mass =
Thus, 12.75g of Sodium Nitrate can be formed