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Supernova

Answer:

4.06 g BaCl₂

Explanation:

Molarity (M) is given by this formula:

  • [tex]\displaystyle \bf{M = \frac{moles \ of \ solute}{Liters\ of\ Solution} = \frac{mol}{L}[/tex]

We are given M and L in this equation; we can solve for moles of BaCl₂ and use stoichiometry to convert this amount to grams.

First convert 50.0 mL to L:

  • 50.0 mL → 0.05 L

Plug this value for L and 0.390 for M into the equation.

  • [tex]\displaystyle \bf{0.390=\frac{mol}{.05}[/tex]

Solve for mol by multiplying .05 to both sides.

  • [tex]\bf{mol = .0195[/tex]

We have .0195 mol BaCl₂. Now let's use stoichiometry to calculate the amount of grams in .0195 mol BaCl₂.

The molar mass of BaCl₂ is 208.23 g so by using stoichiometry...

  • [tex]\displaystyle \bf{.0195 \ mol \ BaCl_2 \cdot \frac{208.23 \ g\ BaCl_2}{1\ mol\ BaCl_2} = 4.060485\ g \ BaCl_2[/tex]  

There are 4.06 g of BaCl₂ in this solution.

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