Chemistry, 28.12.2019 23:31, tagerryawilson6

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

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answered: sanafarghal

Percentage yield of carbonic acid: 88%.

Explanation:

Balance the equation:

${\bf 2}\;\text{NaHCO}_3 \to \text{Na}_2\text{CO}_3 + \text{H}_2\text{CO}_3$ .

Relative atomic mass data from a modern periodic table:

Na: 22.990;H: 1.008;C: 12.011;O: 15.999.

Formula mass:

NaHCO₃: $22.990 + 1.008 + 12.011 + 3\times 15.999=84.006\;\text{g}\cdot\text{mol}^{-1}$ .H₂CO₃: $2\times 1.008+ 12.011 + 15.999 = 62.024\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of NaHCO₃ in the sample:

$\displaystyle n(\text{NaHCO}_3)= \frac{m}{M} = \frac{3.24}{84.006} = 0.0385687\;\text{mol}$ .

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.

$\displaystyle n(\text{H}_2\text{CO}_3) = \frac{1}{2}\;n(\text{NaHCO}_3) = \frac{1}{2} \times 0.0385687 = 0.0192843\;\text{mol}$

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

What will be the theoretical yield of H₂CO₃?

$m(\text{H}_2\text{CO}_3,\;\text{theoretical}) = n\cdot M = 0.0192843\times 62.024 = 1.19609\;\text{g}$ .

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.

$m(\text{H}_2\text{CO}_3,\;\text{actual}) = m(\text{solid},\;\text{initial}) - m(\text{solid},\;\text{final})= 3.24 - 2.19 = 1.05\;\text{g}$ .

What's the percentage yield of H₂CO₃?

$\displaystyle \begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\% \\&= \frac{1.05}{1.19609}\times 100\%\\&=87.8\%\end{aligned}$ .

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.

answered: Guest

1.consider the following balanced equation and answer parts a through. show calculations for each part.

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.

explanation:

answered: Guest

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

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During a laboratory experiment, a 3.24-gram sample of nahco3 was thermally decomposed. in this exper...