m(Ba So4)=120г n(Ba So4)

To calculate the number of moles (\(n\)) of barium sulfate (\(BaSO_4\)), you can use the formula:

\[ n = \frac{m}{M} \]

where:

• \(m\) is the mass of the substance (in grams),
• \(M\) is the molar mass of the substance (in g/mol).

Given:

• \(m(BaSO_4) = 120 \, g\)

First, we need to determine the molar mass of \(BaSO_4\):

• The molar mass of Barium (Ba) = 137.33 g/mol
• The molar mass of Sulfur (S) = 32.07 g/mol
• The molar mass of Oxygen (O) = 16.00 g/mol (and there are 4 oxygen atoms in sulfate)

Calculating the molar mass of \(BaSO_4\):

\[ M(BaSO_4) = M(Ba) + M(S) + 4 \times M(O) \] \[ M(BaSO_4) = 137.33 \, g/mol + 32.07 \, g/mol + 4 \times 16.00 \, g/mol \] \[ M(BaSO_4) = 137.33 \, g/mol + 32.07 \, g/mol + 64.00 \, g/mol \] \[ M(BaSO_4) = 233.40 \, g/mol \]

Now plug in the values to find \(n(BaSO_4)\):

\[ n(BaSO_4) = \frac{120 \, g}{233.40 \, g/mol} \approx 0.514 \, mol \]

So, the number of moles of \(BaSO_4\) is approximately 0.514 moles.