# 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?

Feb 18, 2016

NaI - Sodium Iodide

$P b {\left(N {O}_{3}\right)}_{2} + 2 N a I - \to P b {I}_{2} + 2 N a N {O}_{3}$

Moles of Lead(II) Nitrate = $m a s s / m o l a r m a s s$
= 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
$\frac{1}{2} = \frac{0.075}{x}$
$x = 2 \times 0.075$
$x = 0.150$
Thus that means 0.150 moles of Sodium Nitrate
To calculate mass of Sodium Nitrate
Since we know that Moles = $M a s s / M o l a r M a s s$
Thus Mass = $M o l e s \times M o l a r M a s s$
Molar Mass of $N a N {O}_{3}$
Na - 22.99
N - 14.01
O - 3(16) = 48
85g/mol
Thus, Mass = $0.150 \times 85$
Mass = $12.75$
Thus, 12.75g of Sodium Nitrate can be formed