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

Related questions
See all questions in Stoichiometry
Impact of this question
31214 views around the world
You can reuse this answer
Creative Commons License