Answer:
<h2>Pb(NO3)2- Lead Nitrate</h2><h2>NaI - Sodium Iodide</h2>
<h2>P</h2><h2>b</h2><h2>(</h2><h2>N</h2><h2>O</h2><h2>3</h2><h2>)</h2><h2>2</h2><h2>+</h2><h2>2</h2><h2>N</h2><h2>a</h2><h2>I</h2><h2>−</h2><h2>→</h2><h2>P</h2><h2>b</h2><h2>I</h2><h2>2</h2>
+
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
<h2>0.075 : x</h2>
1
2
=
0.075
x
x
=
2
×
0.075
x
=
0.150
<h2>Thus that means 0.150 moles of Sodium Nitrate</h2><h2>To calculate mass of Sodium Nitrate</h2><h2>Since we know that Moles = </h2>
M
a
s
s
/
M
o
l
a
r
M
a
s
s
<h2>Thus Mass = </h2><h2>M</h2><h2>o</h2><h2>l</h2><h2>e</h2><h2>s</h2><h2>×</h2><h2>M</h2><h2>o</h2><h2>l</h2><h2>a</h2>
r
<h2>M</h2><h2>a</h2><h2>s</h2><h2>s</h2>
<h2>Molar Mass of </h2><h2>N</h2><h2>a</h2><h2>N</h2><h2>O</h2><h2>3</h2>
<h2>Na - 22.99</h2><h2>N - 14.01</h2><h2>O - 3(16) = 48</h2><h2>85g/mol</h2><h2>Thus, Mass = </h2><h2>0.150</h2><h2>×</h2><h2>85</h2>
<h2>Mass = </h2><h2>12.75</h2>
<h2>Thus, 12.75g of Sodium Nitrate can be formed</h2>
