Question

when nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretical yields of each product using stoichiometry if the mass of the nahco3 sample is 3.24 grams. (show work for both) in an actual decomposition of nahco3, the mass of one of the products was measured to be 2.01 grams. identify which product this could be and justify your reasoning. calculate the percent yield of the product identified in part b. (show your work)

Answer 1

Using stoichiometry, 3.24 grams $NaHCO_{3}$ will yield 2.04 grams $Na_{2} CO_{3}$ and 1.20 grams $H_{2} CO_{3}$. But in an actual decomposition where the mass of one of the products was measured to be 2.01 grams, which is $Na_{2} CO_{3}$, the percent yield is 98.34%.

Stoichiometry is defined as the relationship between  the amounts of substances that are involved in chemical reactions.

Mole ratio relates the amounts in moles of any two substances by looking at the coefficients in front of each species in the balanced chemical equation.

For the decomposition of $NaHCO_{3}$, it follows the balanced chemical equation:

$2NaHCO_{3} = Na_{2} CO_{3}+H_{2} CO_{3}$

2 moles of $NaHCO_{3}$ will decompose into 1 mole of $Na_{2} CO_{3}$ and 1 mole of $H_{2} CO_{3}$

mole ratio = 2 : 1 : 1

If the mass of $NaHCO_{3}$ sample is 3.24 grams, then

molar mass = m/n

84.007 g/mol = 3.24 g/n

n = 0.03856821455 mol $NaHCO_{3}$

and using the mole ratio 2 : 1 : 1, theoretically,

0.03856821455 mol $NaHCO_{3}$ will yield 0.01928410728 mol $Na_{2} CO_{3}$ and 0.01928410728 mol $H_{2} CO_{3}$.

If there's 0.01928410728 mol $Na_{2} CO_{3}$, its mass is:

molar mass = m/n

105.9888 g/mol = m/ 0.01928410728

m = 2.043899389 g $Na_{2} CO_{3}$

On the other hand, if there's 0.01928410728 mol $H_{2} CO_{3}$, then

molar mass = m/n

62.03 g/mol = m/0.01928410728

m = 1.196193174 g $H_{2} CO_{3}$

If the mass of one of the products was measured to be 2.01 grams, then it must be     $Na_{2} CO_{3}$ since the experimental value is much closer to the theoretical mass.

Calculating its percent yield,

percent yield = actual yield/theoretical yield x 100%

percent yield = 2.01 g/2.043899389 g x 100%

percent yield = 98.34143552% = 98.34%

To learn more about stoichiometry: brainly.com/question/25829169

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