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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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Explanation:

Hello.

In this case, since the first-order reaction is said to be linearly related to the rate of reaction:

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When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

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