Stoichiometric Calculations

Balance the following equation:

$$NaBr + Ca(OH)_2 \rightarrow CaBr_2 + NaOH$$

Our first step is to count the number of moles of each element on both sides. When elements are in brackets, like (OH) here, we multiply the subscript by each element. Therefore, (OH)₂, is the same as O₂H₂ in terms of numbers.

Let's start with the left side:

Sodium (Na): 1 Bromine (Br): 1 Calcium (Ca): 1 Oxygen (O): 2 Hydrogen (H): 2

Now for the right:

Sodium (Na): 1 Bromine (Br): 2 Calcium (Ca): 1 Oxygen (O): 1 Hydrogen (H):1

So we have three species that aren't balanced: Br, O, and H

Firstly, we can multiply our sodium hydroxide (NaOH) by 2, so that our oxygen and hydrogen atoms are now balanced:

$$NaBr + Ca(OH)_2 \rightarrow CaBr_2 + 2NaOH$$

Now we have a new problem, our sodium atoms are unbalanced! So for our last step, we can multiply the sodium bromide (NaBr) compound by 2, so that our bromine and sodium atoms are all balanced.

$$2NaBr+ Ca(OH)_2 \rightarrow CaBr_2 + 2NaOH$$

Using the balanced equation below, calculate the number of moles of iron (III) chloride (FeCl₃) that are produced when 1.75 moles of chlorine gas (Cl₂) are reacted with enough iron (Fe).

$$2 Fe + 3Cl_2 \rightarrow 2FeCl_3$$

So our first step has been done for us since the equation is already balanced, so onto step 2!

The stoichiometric ratio is simply the ratio of our starting species (Cl₂) to our desired species (FeCl₃). This ratio is given to us by the coefficients.

We multiply our amount of chlorine by this ratio, so that the moles of chlorine cancel, leaving us with moles of FeCl₃:

$$1.75\,mol\,Cl_2*\frac{2\,mol\,FeCl_3}{3\,mol\,Cl_2}$$

$$1.75\cancel{mol\,{Cl_2}}*\frac{2\,mol\,FeCl_3}{3\cancel{mol\,Cl_2}}$$

$$1.75*\frac{2}{3}\,mol\,FeCl_3=1.17\,mol\,FeCl_3$$

Using the balanced equation below, calculate how many grams of Carbon dioxide (CO₂) are produced when 54.6 g of ethane (C₂H₆) are combusted with enough oxygen gas (O₂) to react

$$2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6 H_2O$$

• Atomic mass (C): 12.01 g/mol Atomic mass (H): 1.01 g/mol
• Atomic mass (O): 16.00 g/mol

For our first step, we need to convert grams of ethane to moles. To do this, we first need to calculate the compound's molar mass:

$$2(12.01\frac{g}{mol})+6(1.01\frac{g}{mol})=30.08\frac{g}{mol}$$

Next, we divide the mass by the molar mass, so that the grams units cancel, and we are left in units of moles:

$$\frac{54.6\cancel{\,g}\,C_2H_6}{\frac{30.08\cancel{\,g}\,C_2H_6}{mol}}=1.82\,mol\,C_2H_6$$

Next, we multiply by the ratio of ethane to Carbon dioxide:

$$1.82\cancel{\,mol\,C_2H_6}*\frac{4\,mol\,CO_2}{2\cancel{\,mol\,C_2H_6}}=3.64\,mol\,CO_2$$

Now we need to convert back to moles, to do this, we multiply by the molar mass. But first, we need to calculate the molar mass of CO₂:

$$12.01\frac{g}{mol}+2(16.00\frac{g}{mol})=44.01\frac{g}{mol}$$

Finally, we can multiply the number of moles by the molar mass to get the gram amount:

$$3.64\cancel{\,mol}\,CO_2*44.01\frac{g}{\cancel{mol}}\,CO_2=160.2\,g\,CO_2$$