Question

How many grams of nitric acid will react completely with a block of iron metal that is 4.5 cm by 3.0 cm by 3.5 cm, if the density of iron is 7.87 g/mL? Show all steps of your calculation as well as the final answer. Fe + HNO3 → Fe(NO3)3 + H2

Answer 1

Answer:

1258 g HNO3

Explanation:

First, we need to calculate the grams of Fe we have using the density and the volume given. For this, use the definition of density which is:

Density = mass / volume

So

Mass = Density x volume

For the statement, the volume of Fe will be:

Volume = (4.5 cm)(3.0 cm)(3.5 cm) = 47.2 cm^3

And as 1 cm^3 is equal to 1 mL, then:

Volume = 47.2 mL

So the mass of Fe we have is:

Mass = (7.87 g/mL)(47.2 mL) = 371.9 g Fe

To find the amount of nitic acid that reacts we need to work with moles, then:

Moles of Fe = $371.9gFe (\frac{1 mole Fe}{55.85 gFe})=6.658 moles Fe$

And as the balanced reaction is  2Fe + 6HNO3 → 2Fe(NO3)3 + 3H2, then the moles of nitric acid that reacts with the given amount of iron is:

Moles of HNO3 = $6.658moleFe (\frac{6 moles HNO3}{2 moles Fe} )=19.97 moles HNO3$

But they are asking for the grams that react, so we need the molecular mass of HNO3 to do the conversion

Molecular mass of HNO3 = 1.008 g H + 14.01 g N + 3(16.00 g O) = 63.02 g HNO3

Finally, the mass of HNO3 that reacts with the amount of iron given is:

mass HNO3 = $19.97molesHNO3 (\frac{63.02 g HNO3}{1 mole HNO3} ) = 1258 g HNO3$

Answer 2

Balance the equation first: 

2 Fe+6 HNO3→2 Fe(NO3)3+3H2

Then calculate mass of Iron :

4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe

Now use Stoichiometry:

371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3

Convert moles of nitric acid to grams

19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3