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Intro to Stoichiometry

Stoichiometry allows us to predict the quantities of products or reagents across a chemical reaction.

Typically this is done in a few steps:
  1. Calculating the number of moles of the reagents
  2. Calculating the number of moles of the products using the stoichiometric ratio
  3. Calculating the mass or pressure of the products.

Moles are the central unit!

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In addition we can convert between mass and volume of a pure substance by looking at it's density,
δ=m/V\boxed{\delta=m/V}

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Stoichiometry of a Reaction

We use the coefficients of the balanced reaction along with the our equations that convert mass, volume, and concentration into moles to predict the quantities of reactants and products in a chemical reaction.

To answer any stoichiometry problem, focus on converting to and from moles!

General Steps to Solving a Stoichiometry Problem:

  1. Convert the values given in the problem about a reactant or product to a number of moles
  2. Use the stoichiometric coefficients from the balanced reaction to find the number of moles of the unknown you are being asked for
  3. Convert the number of moles of your unknown to a mass, or whatever quantity you are being asked for


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Example

2.6g of sodium metal (Na) reacts with water to form NaOH and H2 according to the unbalanced reaction below. Once the reaction is complete how many grams of NaOH are formed?
Na(s)+  H2O(g)  NaOH(s)+  H2(g)Na_{(s)}+\ \ H_2O_{(g)}\rightarrow\ \ NaOH_{(s)}+\ \ H_{2(g)}



Wize Tip
On an exam, your prof will NOT specify if an equation is balanced or unbalanced. Always double check that the equation is balanced. If it's not, balance the equation before continuing!
The equation must be balanced in order to get the correct answer!


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Step 1: Balance the reaction
2 Na(s)+2 H2O(g)2 NaOH(s)+H2(g)2\ Na_{(s)}+2\ H_2O_{(g)}\rightarrow 2\ NaOH_{(s)}+H_{2(g)}


Step 2: Find the number of moles of sodium
nNa=mMW=2.6 g22.990 g/mol=0.113 moln_{Na}=\dfrac{m}{MW}=\dfrac{2.6\ g}{22.990\ g/mol}=0.113\ mol


Step 3: Find the number of moles of NaOH
nNaOH=nNa×coefficient NaOHcoefficient Na=0.113 mol×22=0.113 moln_{NaOH}=n_{Na}\times \dfrac{coefficient\ NaOH}{coefficient\ Na}=0.113\ mol\times \dfrac{2}{2}=0.113\ mol


Step 4: Convert the number of moles of NaOH to a mass

n=mMn=\frac{m}{M}
mNaOH=n×MW=0.113 mol×39.997 g/mol=4.520 g=4.5 gm_{NaOH}=n\times MW=0.113\ mol \times 39.997\ g/mol=4.520\ g = 4.5\ g

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Example: Converting Mass of Reactant to Mass of Product

Calculate the mass of hydrogen gas produced if 0.550 g of iron powder is reacted with an excess amount of sulfuric acid.

 Fe (s)   +   H2SO4 (aq)        Fe2(SO4)3 (aq)   +   H2 (g)\ Fe\ \left(s\right)\ \ \ +\ \ \ H_2SO_4\ \left(aq\right)\ \ \ \ \rightarrow\ \ \ \ Fe_2\left(SO_4\right)_3\ \left(aq\right)\ \ \ +\ \ \ H_2\ \left(g\right)


Step 1-Balance the equation
 2Fe (s)   +   3H2SO4 (aq)        Fe2(SO4)3 (aq)   +   3H2 (g)\ 2Fe\ \left(s\right)\ \ \ +\ \ \ 3H_2SO_4\ \left(aq\right)\ \ \ \ \rightarrow\ \ \ \ Fe_2\left(SO_4\right)_3\ \left(aq\right)\ \ \ +\ \ \ 3H_2\ \left(g\right)


Step 2- Calculate the moles of iron (Fe(s))
n=m/M
n=0.550g/55.845g/mol
n=0.00985 mol


Step 3-Using the stoichiometric coefficients from the balanced chemical reaction, determine the number of moles of H2(g)

0.00985 mol Fe x 3 moles of H22 moles Fe=moles of H2(g)0.00985\ mol\ Fe\ x\ \frac{3\ moles\ of\ H_2}{2\ moles\ Fe}=moles\ of\ H_2\left(g\right)
moles H2(g)=0.01477 moles


Step 4-Calculate the mass of hydrogen gas
n=m/M
m=nM
m=(0.01477moles)(2.02g/mol)
m=0.0298g
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Example: Solution Stoichiometry to Determine Mass of Product

Lithium metal was added to a 25mL of a 1.3M solution of Ag2SO4. The unbalanced chemical reaction is shown below.
Ag2SO4(aq)+Li(s)Li2SO4(aq)+Ag(s)Ag_2SO_{4(aq)}+Li_{(s)}\rightarrow Li_2SO_{4(aq)}+Ag_{(s)}
Once the reaction has gone to completion, what mass of silver metal is produced?

a) 0.0070 g
b) 7.0 g
c) 3.5 g
d) 0.0035 g

1. Balance the reaction
Ag2SO4(aq)+2 Li(s)Li2SO4(aq)+2 Ag(s)Ag_2SO_{4(aq)}+2\ Li_{(s)}\rightarrow Li_2SO_{4(aq)}+2\ Ag_{(s)}

2. Find the number of moles of Ag2SO4 originally in solution
  • Recall n=cv
  • c must be in mol/L =M
  • V must be in L
  • Volume here is 25mL
  • 1000mL=1L
  • 25mL=0.025L
nAg2SO4=CV=(1.3 mol/L)(0.025L)=0.0325 moln_{Ag_2SO_4}=CV=(1.3\ mol/L)\left(0.025L\right)=0.0325\ mol

3. Use the stoichiometric coefficients to find the number of moles of Ag produced
nAg=nAg2SO4×21=0.065 moln_{Ag}=n_{Ag_2SO_4}\times \dfrac{2}{1}=0.065\ mol

4. Find the mass of silver from the moles
n=m/M
Rearrange to solve for m.
m=n×MW=(0.065 mol)(107.87 g/mol)=7.01 gm=n\times MW=(0.065\ mol)(107.87\ g/mol)=7.01\ g

Answer: B

Practice: Calculate the Mass of Product

16.0 mL of Hg (mp = –38.8oC, density = 13.56 g/mL) reacts with fluorine gas and the reaction goes to completion as shown below.
Hg(l)+F2(g)HgF2(s)Hg_{(l)}+F_{2(g)}\rightarrow HgF_{2(s)}

Once the reaction has gone to completion, what mass of mercury fluoride, HgF2 is produced?

Practice: Stoichiometry Calculations Application

A solution that is 183mM (millimolar) NaCl(aq) is isoosmotic with plasma. This means that cells don't swell or shrink when in this solution. How many grams of sodium chloride are required to make 150mL of isoosmotic NaCl(aq)? The molar mass of NaCl is 58.44g/mol.
Extra Practice