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02 question (4 points) how many atoms of iron are there in 2.01 moles o…

Question

02 question (4 points)
how many atoms of iron are there in 2.01 moles of each of the following?
1st attempt
part 1 (1 point)
wolframite, fewo₄ atoms fe
part 2 (1 point)
pyrite, fes₂ atoms fe
part 3 (1 point)
magnetite, fe₃o₄ atoms fe
part 4 (1 point)
hematite, fe₂o₃ atoms fe

Explanation:

Part 1: For \(FeWO_4\)

Step1: Determine moles of Fe

In \(FeWO_4\), there is 1 mole of Fe per mole of \(FeWO_4\). So moles of Fe = 2.01 mol.

Step2: Calculate number of atoms

Using Avogadro's number \(N = n\times N_A\), where \(n = 2.01\) mol and \(N_A=6.022\times 10^{23}\) atoms/mol.
\(N=2.01\times6.022\times 10^{23}=1.21\times 10^{24}\) atoms.

Part 2: For \(FeS_2\)

Step1: Determine moles of Fe

In \(FeS_2\), there is 1 mole of Fe per mole of \(FeS_2\). So moles of Fe = 2.01 mol.

Step2: Calculate number of atoms

Using \(N = n\times N_A\), with \(n = 2.01\) mol and \(N_A = 6.022\times 10^{23}\) atoms/mol.
\(N=2.01\times6.022\times 10^{23}=1.21\times 10^{24}\) atoms.

Part 3: For \(Fe_3O_4\)

Step1: Determine moles of Fe

In \(Fe_3O_4\), there are 3 moles of Fe per mole of \(Fe_3O_4\). So moles of Fe \(=2.01\times3 = 6.03\) mol.

Step2: Calculate number of atoms

Using \(N=n\times N_A\), with \(n = 6.03\) mol and \(N_A=6.022\times 10^{23}\) atoms/mol.
\(N = 6.03\times6.022\times 10^{23}=3.63\times 10^{24}\) atoms.

Part 4: For \(Fe_2O_3\)

Step1: Determine moles of Fe

In \(Fe_2O_3\), there are 2 moles of Fe per mole of \(Fe_2O_3\). So moles of Fe \(=2.01\times 2=4.02\) mol.

Step2: Calculate number of atoms

Using \(N=n\times N_A\), with \(n = 4.02\) mol and \(N_A=6.022\times 10^{23}\) atoms/mol.
\(N=4.02\times6.022\times 10^{23}=2.42\times 10^{24}\) atoms.

Answer:

  • Part 1: \(1.21\times 10^{24}\) atoms Fe
  • Part 2: \(1.21\times 10^{24}\) atoms Fe
  • Part 3: \(3.63\times 10^{24}\) atoms Fe
  • Part 4: \(2.42\times 10^{24}\) atoms Fe