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marusya05 [52]
3 years ago
6

If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?

Chemistry
2 answers:
kap26 [50]3 years ago
6 0
Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
  1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3     1mol CO2
---------------------- x -------------------------
            1                  84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!
alekssr [168]3 years ago
3 0

Answer : The number of moles of carbon dioxide is, 0.156 mole

Explanation : Given,

Mass of citric acid = 10 g

Molar mass of citric acid = 192 g/mole

The balanced chemical reaction will be,

C_6H_8O_7+3NaHCO_3\rightarrow Na_3C_6H_5O_7+3CO_2+3H_2O

From the balanced chemical reaction, we conclude that 1 mole of C_6H_8O_7 react with 3 moles of NaHCO_3 to give 1 mole of Na_3C_6H_5O_7, 3 moles of carbon dioxide gas and 3 moles of water as a product.

First we have to calculate the moles of citric acid.

\text{Moles of citric acid}=\frac{\text{Mass of citric acid}}{\text{Molar mass of citric acid}}=\frac{10g}{192g/mole}=0.052mole

Now we have to calculate the moles of carbon dioxide.

As, 1 mole of citric acid react to gives 3 moles of carbon dioxide

So, 0.052 mole of citric react to gives \frac{3}{1}\times 0.052=0.156 moles of carbon dioxide

Therefore, the number of moles of carbon dioxide is, 0.156 mole

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rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

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