Question

a chemist adds 370.0 ml of a 1.7 mol/l iron(iii) bromide (febr₃) solution to a reaction flask. calculate the mass in kilograms of iron(iii) bromide the chemist has added to the flask. be sure your answer has the correct number of significant digits. kg

Explanation:

Step1: Calculate the number of moles of \(FeBr_3\)

Use the formula \(n = C\times V\), where \(C\) is the concentration and \(V\) is the volume.
Given \(C = 1.7\space mol/L\) and \(V=370.0\space mL = 0.3700\space L\)
\(n = 1.7\space mol/L\times0.3700\space L=0.629\space mol\)

Step2: Calculate the molar mass of \(FeBr_3\)

The molar mass of \(Fe = 55.85\space g/mol\), \(Br = 79.90\space g/mol\)
\(M(FeBr_3)=55.85+(3\times79.90)=55.85 + 239.7=295.55\space g/mol\)

Step3: Calculate the mass of \(FeBr_3\) in grams

Use the formula \(m=n\times M\)
\(m = 0.629\space mol\times295.55\space g/mol\approx186.9\space g\)

Step4: Convert grams to kilograms

Since \(1\space kg = 1000\space g\), \(m=\frac{186.9\space g}{1000}=0.1869\space kg\)

Rounding to two significant digits (because \(1.7\) has two significant digits), \(m = 0.19\space kg\)

Answer:

\(0.19\space kg\)