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the value of calculating a mole


An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy’s density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm3. He accidently breaks off a 1.203 cm3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with teh copper, how many grams of H2(g) are released into the atmosphere because of the amateur’s carelessness? (Note that the situation is fiction.)


Step 1: Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H+(aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe2+(aq).

\[Fe_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + Fe^{2+}_{(aq)}\]

Step 2: Write down all the given information

Alloy density = (3.15g alloy/ 1L alloy)

x grams of alloy = 45% copper = (45g Cu(s)/100g alloy)

x grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)

1 liter alloy = 1000cm3 alloy

alloy sample = 1.203cm3 alloy

Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed.

Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound.

Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted.

1.203cm3 alloy(1liter alloy/1000cm3 alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10-5 mol Fe(s)

Make sure all the units cancel out to give you moles of Fe(s). The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass.

The balanced equation must now be used to convert moles of Fe(s) to moles of H2(g). Remember that the balanced equation’s coeffiecients state the stoichiometric factor or mole ratio of reactants and products.

3.74 x 10-5 mol Fe (s) (1mol H2(g)/1mol Fe(s)) = 3.74 x 10-5 mol H2(g)

Step 5: Check units

The question asks for how many grams of H2(g) were released so the moles of H2(g) must still be converted to grams using the molar mass of H2(g). Since there are two H in each H2, its molar mass is twice that of a single H atom.

molar mass = 2(1.00794g/mol) = 2.01588g/mol

3.74 x 10-5 mol H2(g) (2.01588g H2(g)/1mol H2 (g)) = 7.53 x 10-5 g H2(g) released