The problem on my homework is:
What mass in grams of iron metal is required to react with 2.50 L of .1500 M of sulfuric acid to produce 18.35g of iron (III) sulfate?
please help!
To solve this problem, we need to use stoichiometry and the balanced chemical equation.
Step 1: Write the balanced chemical equation for the reaction between iron and sulfuric acid.
2 Fe + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O
Step 2: Calculate the number of moles of sulfuric acid.
Using the given concentration and volume, we can first find the number of moles of sulfuric acid:
Moles of sulfuric acid = concentration x volume
= 0.1500 M x 2.50 L
= 0.375 moles H2SO4
Step 3: Determine the ratio between H2SO4 and Fe.
From the balanced equation, we can see that the ratio between H2SO4 and Fe is 3:2. That means for every 3 moles of H2SO4, we need 2 moles of Fe.
Step 4: Calculate the number of moles of Fe.
Using the mole ratio, we can determine the number of moles of Fe required:
Moles of Fe = (moles of H2SO4 x 2 moles of Fe) / 3 moles of H2SO4
= (0.375 moles x 2) / 3
= 0.250 moles of Fe
Step 5: Convert moles of Fe to grams.
Finally, we can convert the moles of Fe to grams using the molar mass of iron:
Mass of Fe = moles of Fe x molar mass of Fe
= 0.250 moles x 55.845 g/mol
= 13.96 grams
Therefore, approximately 13.96 grams of iron metal is required to react with 2.50 L of 0.1500 M sulfuric acid to produce 18.35 grams of iron (III) sulfate.