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3 years ago

Determine whether a precipitate will form when 0.96g of Na2CO3 is combined with 0.20g BaBr2 in a 10 L solution.

Chemistry

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

Qsp > Ksp, BaCO3 will precipitate

Explanation:

The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M

Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles

concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M

Hence;

[Ba^2+] = 6.7 * 10^-5 M

[CO3^2-] = 9.1 * 10^-4 M

Qsp = [6.7 * 10^-5] [9.1 * 10^-4]

Qsp = 6.1 * 10^-8

But, Ksp for BaCO3 is 5.1*10^-9.

Since Qsp > Ksp, BaCO3 will precipitate

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Explanation:

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Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

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So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

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Hence, the percent yield of the water is 31.98 %

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