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mrs_skeptik [129]
3 years ago
10

If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, what is the limiting reactant?

Chemistry
1 answer:
Georgia [21]3 years ago
4 0

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>
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Answer:

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