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Alexxandr [17]
3 years ago
5

Write net ionic equations for gas-forming reactions. Write a net ionic equation for the reaction that occurs when solid barium c

arbonate is combined with excess aqueous hydroiodic acid.
Chemistry
1 answer:
AysviL [449]3 years ago
7 0

Answer: BaCO3(s) + 2H+(aq) --> Ba+2 + H2O(l) + CO2(g)

Explanation: Net ionic equations show only the species or atoms that participate in the reaction, in this case gas-forming

When solid barium carbonate is combined with excess aqueous hydroiodic acid the reaction that take place is.

BaCO3(s) + 2HI(aq) --> BaI2 (aq) + H2O(l) + CO2(g)

To write a ionic equation is necesary to separte the species that are dissociated in water (ag) and expresed it as ions showing the charge

BaCO3(s) + 2I-(aq) + 2H+(aq) --> Ba+2 + 2I- (aq) + H2O(l) + CO2(g)

Since I- is present in both sides of the reaction, it's eliminated in the net ionic equation, and the final expression is:

BaCO3(s) + 2H+(aq) --> Ba+2 + H2O(l) + CO2(g)

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. KMnO4 oxidizes ethyl benzene to benzoic acid. If this reaction generally results in a 40 % experimental yield, how many grams
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Answer:

530.835 g

Explanation:

First we convert 244 g of benzoic acid (C₇H₆O₂) to moles, using its molar mass:

  • 244 g benzoic acid ÷ 122.12 g/mol = 2.00 moles benzoic acid

Theoretically,<em> one mol of ethyl benzene would produce one mol of benzoic acid</em>. But the experimental yield tells us that one mol of ethyl benzene will produce only 0.4 moles of benzoic acid.

With the above information in mind we convert 2.00 moles of benzoic acid into moles of ethyl benzene:

  • 2.00 moles benzoic acid * \frac{1molEthylBenzene}{0.4molBenzoicAcid} = 5.00 moles ethyl benzene

Finally we <u>convert moles of ethyl benzene </u>(C₈H₁₀)<u> into grams</u>, using its <em>molar mass</em>:

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3 years ago
A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve
mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

                  PV = nRT

or,           n = \frac{PV}{RT}

                 = \frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}

                  = 0.0052 mol

Also,  No. of moles = \frac{mass}{\text{molar mass}}

               0.0052 mol = \frac{mass}{2 g/mol}

                  mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at 25^{o}C

                P_{\text{water vapor}} = 24 mm Hg

                                = \frac{24}{760} atm

                                = 0.03158 atm

Now,   P = \frac{756}{760} - 0.03158

              = 0.963 atm

Hence,   n = \frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}

                 = 0.0056 mol

So, mass of H_{2} = 0.0056 mol × 2

                         = 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

      Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

                              = \frac{0.01013 g}{0.0104 g} \times 100            

                              = 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

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