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a chemist adds 0.30 l of a 0.46 m barium chlorate (ba(clo₃)₂) solution …

Question

a chemist adds 0.30 l of a 0.46 m barium chlorate (ba(clo₃)₂) solution to a reaction flask. calculate the mass in grams of barium chlorate the chemist has added to the flask. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate the number of moles

Use the formula \(n = C\times V\), where \(C = 0.46\space M\) (molarity) and \(V=0.30\space L\) (volume).
\(n=\frac{0.46\space mol}{L}\times0.30\space L = 0.138\space mol\)

Step2: Calculate the molar mass of \(Ba(ClO_3)_2\)

The molar mass of \(Ba\) is \(137.33\space g/mol\), \(Cl\) is \(35.45\space g/mol\), and \(O\) is \(16.00\space g/mol\).
\(M = 137.33+(35.45 + 3\times16)\times2\)
\(M=137.33+(35.45 + 48)\times2\)
\(M=137.33 + 83.45\times2\)
\(M=137.33+166.9\)
\(M = 304.23\space g/mol\)

Step3: Calculate the mass

Use the formula \(m=n\times M\), where \(n = 0.138\space mol\) and \(M = 304.23\space g/mol\)
\(m=0.138\space mol\times304.23\space g/mol\)
\(m = 42.0\space g\) (rounded to three significant digits)

Answer:

\(42.0\space g\)