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
Ayako_Sho
Feb 18, 2016

Pb(NO3)2- Lead Nitrate
NaI - Sodium Iodide

#Pb(NO_3)_2 + 2NaI --> PbI_2+ 2NaNO_3#

Moles of Lead(II) Nitrate = #mass // molar mass#
= 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
#1/2 = 0.075/x#
#x = 2 xx 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 = #Mass//Molar Mass#
Thus Mass = #Mol es xx Molar Mass#
Molar Mass of #NaNO_3#
Na - 22.99
N - 14.01
O - 3(16) = 48
85g/mol
Thus, Mass = #0.150 xx 85#
Mass = #12.75#
Thus, 12.75g of Sodium Nitrate can be formed

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