Using balanced equations
Balanced equations are used to allow chemists to calculate how much product they will produce from their reactants. The co-efficients in balanced equations represent the number of moles reacting and being produced, it is rare that this exact ratio will be used in a reaction.
For example, some reactants can be very expensive so the experiment may be scaled down, usually one reactant is in excess or perhaps the chemist knows approximately how much product they need.
Calculating mass
Being able to calculate masses based on balanced equations is very important. Moles can be compared but masses cannot so the following equation is used to help convert between moles and mass:
\(Number\ of \ moles = \frac{mass}{formula\ mass}\)
Example one
Calculate the mass of sodium chloride produced when 11.5 g of sodium is reacted with excess chlorine gas.
Step one
Start with the balanced equation:
\( 2Na + Cl_{2} \rightarrow 2NaCl\)
Step two
Use the information from the question to calculate how many moles of reactant are being used:
\(moles \ of \ sodium = \frac{11.5}{23} = 0.5\)
Step three
Use the balanced equation to work out how many moles of product this equates to using mole ratios.
\(2 \ moles \ of \ Na \leftarrow \rightarrow 2 \ moles\ of \ NaCl\)
\(1 \ mole \ of \ Na \leftarrow \rightarrow 1 \ mole\ of \ NaCl\)
\(0.5 \ moles \ of \ Na \leftarrow \rightarrow 0.5 \ moles\ of \ NaCl\)
Step four
Calculate the mass of product produced using the moles calculated in step 3 and the gram formula mass of the product:
\( gfm \ NaCl=23 + 35.5 = 58.5 \)
\( mass \ sodium \ chloride = moles \times gfm = 0.5 \times 58.5 = 29.25g \)
Example two
Calculate the mass of oxygen required to react with iron to produce 240 g of iron (III) oxide.
Step one
Start with the balanced equation:
\(4Fe+3O_{2}\rightarrow2Fe_{2}O_{3}\)
Step two
Use the information in the question and the gram formula mass to calculate the moles of product produced:
\( gfm \ of \ iron \ (III) \ oxide \ [Fe_{2}O_{3}] = (2\times56) + (3\times16) = 112 + 48 = 160\)
\(moles \ of \ iron \ (III) \ oxide = \frac{mass}{gfm}=\frac{240}{160} = 1.5 \ moles\)
Step three
Use the balanced equation to work out how many moles of reactant this equates to using mole ratios:
\(3 \ moles \ O_{2} \leftarrow \rightarrow 2 \ moles \ Fe_{2}O_{3}\)
\( 1.5 \ moles \ O_{2} \leftarrow \rightarrow 1 \ mole \ Fe_{2}O_{3}\)
\( 2.25 \ moles \ O_{2} \leftarrow \rightarrow 1.5 \ moles \ Fe_{2}O_{3} \)
Step four
Calculate the mass of reactant needed using the moles calculated in step 3 and the gram formula mass of the reactant.
\(gfm \ O_{2} = 2 \times 16 = 32\)
\( mass = moles \times gfm = 2.25 \times 32 = 72g\)