Question

a sample of 8.34 moles of iron(iii) sulfite will weigh how many grams? 1.27e3 g 2.93e3 g 0.0237 g 3.34e3 g

Explanation:

Step1: Determine the formula of iron(III) sulfite

Iron(III) has a charge of \(Fe^{3+}\), sulfite is \(SO_{3}^{2 -}\). Using the criss - cross method, the formula is \(Fe_{2}(SO_{3})_{3}\).

Step2: Calculate the molar mass of \(Fe_{2}(SO_{3})_{3}\)

• Molar mass of \(Fe\): \(55.85\space g/mol\), for \(2\space Fe\) atoms: \(2\times55.85 = 111.7\space g/mol\)
• Molar mass of \(S\): \(32.07\space g/mol\), for \(3\space S\) atoms: \(3\times32.07=96.21\space g/mol\)
• Molar mass of \(O\): \(16.00\space g/mol\), for \(9\space O\) atoms: \(9\times16.00 = 144.0\space g/mol\)
• Molar mass of \(Fe_{2}(SO_{3})_{3}\): \(M=111.7+96.21 + 144.0=351.91\space g/mol\)

Step3: Use the formula \(m=n\times M\)

Given \(n = 8.34\space mol\) and \(M = 351.91\space g/mol\)
\(m=8.34\times351.91\space g\)
\(m = 8.34\times351.91=8.34\times(350 + 1.91)=8.34\times350+8.34\times1.91\)
\(8.34\times350=(8 + 0.34)\times350=8\times350+0.34\times350=2800+119 = 2919\)
\(8.34\times1.91=(8+0.34)\times(2 - 0.09)=16-0.72+0.68 - 0.0306=15.9294\)
\(m=2919+15.9294\approx2935\space g\approx2.93\times 10^{3}\space g\)

Answer:

\(2.93E3\space g\)