How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 percent by mass iron(II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem.
2 Al (s) + 3 Fe(NO3)2 (aq) 3 Fe (s) + 2 Al(NO3)2 (aq) (4 points)
You need to learn how to write arrows. --> or ==> or >>> will do it. Technically, the equation you wrote was incorrect since it doesn't separate the reactants from the products.
305 g sample Fe(NO3)2 has how much Fe(NO3)2? That's 305 x 0.755 = ?
mols Fe(NO3)2 = grams/molar mass
Use the coefficients in the balanced equation to convert mols Fe(NO3)2 to mols Fe.
Now convert mols Fe to grams Fe. g = mols x atomic mass Fe.
To determine the number of grams of iron metal produced, we need to follow a series of steps.
Step 1: Calculate the number of moles of iron(II) nitrate (Fe(NO3)2) present in the solution.
First, we convert the given mass of the iron(II) nitrate solution to moles. To do this, we need to know the molar mass of Fe(NO3)2.
1 mole of Fe(NO3)2 = 55.85 g + 2(14.01 g) + 6(16.00 g) = 179.85 g
Now, we can calculate the number of moles of Fe(NO3)2:
Number of moles = (Mass of the solution / Molar mass of Fe(NO3)2)
Number of moles = (305 g / 179.85 g/mol)
Step 2: Calculate the number of moles of iron (Fe) produced.
From the balanced chemical equation, we can see that the stoichiometric ratio between Fe(NO3)2 and Fe is 3:3, or 1:1. This means that for every mole of Fe(NO3)2, we will produce one mole of Fe.
Number of moles of Fe = Number of moles of Fe(NO3)2
Step 3: Convert moles of Fe to grams of Fe.
To convert moles to grams, we need to multiply the number of moles of Fe by its molar mass.
The molar mass of Fe is 55.85 g/mol.
Mass of Fe = Number of moles of Fe x Molar mass of Fe
Step 4: Plug in the values and solve.
Mass of Fe = Number of moles of Fe x Molar mass of Fe
Mass of Fe = (Number of moles of Fe(NO3)2) x (Molar mass of Fe)
Now, let's calculate the answer using the given values.
Step 1:
Number of moles of Fe(NO3)2 = (305 g / 179.85 g/mol) ≈ 1.695 moles
Step 2:
Number of moles of Fe = 1.695 moles
Step 3:
Mass of Fe = 1.695 moles x 55.85 g/mol ≈ 94.54 grams
Therefore, we expect approximately 94.54 grams of iron metal to be produced when 305 grams of a 75.5 percent by mass iron(II) nitrate solution reacts with excess aluminum metal.