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UkoKoshka [18]
2 years ago
11

Excess silver(I) nitrate was added to a 8.500 g mixture containing some amount of barium chloride, and 7.123 g of silver chlorid

e was obtained. The unbalanced equation is
AgNO3 + BaCl2(aq) --> Ba(NO3)2(aq) + AgCl(s).
What is the mass% of BaCl2 in the mixture?
Chemistry
1 answer:
dsp732 years ago
8 0

Answer:

60.88%

Explanation:

The balanced equation of the reaction is given as;

2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)

Since AgNO3 is in excess, the limiting reactant is BaCl2. From the reaction;

1 mol of BaCl2 produces 2 mol of AgCl

Converting to masses;

Mass = Number of mol * Molar mass

BaCl2;

Mass = 1 * 208.23 g/mol = 208.23 g

AgCl;

Mass = 2 * 143.32 g/mol = 286.64 g

208.23 g BaCl2 produces 286.64 g of AgCl

x g BaCl2 produces 7.123 g of AgCl

Solving for x;

x = 7.123 * 208.23 / 286.64 = 5.1745 g

Mass percent = Mass / Total mass of Mixture  * 100

Mass Percent = 5.1745 / 8.500 = 0.6088 * 100 = 60.88%

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ioda

Answer:

Explanation:

From the statement of the problem,

  B₂S₃_{s} + H₂O_{l}  →  H₃BO₃_{aq} + H₂S_{g}

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We that the above expression does not conform with the law of conservation of mass:

To obey the law, we need to derive a balanced reaction equation:

   Let us use the mathematical method to obtain a balanced equation.

let the balanced equation be:

                        aB₂S₃ + bH₂O  →  cH₃BO₃ + dH₂S

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  Conservating B: 2a = c

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                          H: 2b = 3c + 2d

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      b = 6,

      2d = 2(6) - 3(2) = 6, d = 3

Now we can input this into our equation:

                     B₂S₃ + 6H₂O  →  2H₃BO₃ + 3H₂S

    B₂S₃_{s} + 6H₂O_{l}  →  2H₃BO₃_{aq} + 3H₂S_{g}

4 0
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