During a laboratory experiment, a 3.24-gram sample of nahco3 was thermally decomposed. in this experiment, carbon dioxide and water vapors escape and are combined to form carbonic acid. after decomposition, the sample weighed 2.19 grams. calculate the percentage yield of carbonic acid for the reaction. describe the calculation process in detail. nahco3 → na2co3 + h2co3

Percentage yield of carbonic acid: 88%.

Explanation:

Balance the equation:

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Relative atomic mass data from a modern periodic table:

Formula mass:

Number of moles of NaHCO₃ in the sample:

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How many moles of carbonic acid will that many NaHCO₃ produce in theory? The ratio between the coefficient in front of NaHCO₃ and H₂CO₃ in the balanced equation is 2:1. In other words, it takes two moles of NaHCO₃ to produce one mole of H₂CO₃.

In other words, 0.0385687 moles of NaHCO₃ will produce 0.0192843 moles of H₂CO₃.

What will be the theoretical yield of H₂CO₃?

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What's the actual yield of H₂CO₃?

Mass conserves. As a result, the difference between the mass of the solid before and after the reaction is the mass of the H₂O and CO₂ that escaped. The two species combine to form H₂CO₃. The mass of the H₂CO₃ produced will be the same as the change in the mass of the solid.

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What's the percentage yield of H₂CO₃?

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Note: all values in this question are in 3 sig. fig. Keep more than 3 sig. fig. in the calculations but round the final result to 3 sig. fig.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :