Stoichiometry: yield of reaction between hydrochloric acid and calcium carbonate

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  • #1
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Thank you in advance for the correction of the resolution.

Homework Statement



By reacting 225 g of hydrochloric acid with 300 g of calcium carbonate with 80% of purity, 34 L (litres) of gas were obtained, measured at 37oC and a pression of 0.82 atm. What is the yield of this reaction and what mass of calcium carbonate reacted? (Remember that H2CO3, in water, decomposes as water and CO2)

[tex]2HCl_{(aq)} + CaCO_3_{(s)} \rightarrow CaCl_2_{(s)} + H_2CO_3_{(aq)}[/tex]

R = 0.082 (atm*L)/(mol*K).

Homework Equations



[tex]PV = nRT[/tex]

The Attempt at a Solution



[tex]2HCl_{(aq)} + CaCO_3_{(s)} \rightarrow CaCl_2_{(s)} + H_2O_{(aq)} + CO_2_{(g)}[/tex]
Molecular masses: 2 HCl = 2(36.5) = 73 g; 1CaCO3 = 100 g.
Assuming that the expression "with 80% of purity" refers to calcium carbonate (we are not sure):
The molar relation in order to know the limiting reagent (but the purity of the calcium carbonate is 80%, so 300 g * 80% = 240 g):
[tex]1 mol HCl \rightarrow 1 mol CaCO_3[/tex]
[tex]73 g \rightarrow 100g[/tex]
[tex]x \rightarrow 300 \times 0.8[/tex]
x = 175.2 g of HCl. Smaller than the available quantity, 225 g, thus it is in excess. The limiting reagent is CaCO3.
Now, the molar relation between calcium carbonate and the gas, CO2, in order to find how many litres of the gas (y) are produced with 100% yield.
But first, to find the molar volume:
37oC = 37 + 273 = 310 K.
[tex]PV_m = nRT \rightarrow 0.82V_m = 0.082 \times 310 \rightarrow V_m = 31 L[/tex]
[tex]1 mol CaCO_3 \rightarrow 1 mol CO_2[/tex]
[tex]100 g \rightarrow 31L[/tex]
[tex]240 g \rightarrow y[/tex]
y = (31 * 240) / 100
y = 74.4 L.
What is the yield of this reaction?
Since the volume of gas produced with the current yield is 34 L, the yield (R) is:
[tex]R = \frac{34}{74.4}[/tex]
R = 45.6%.
What mass of calcium carbonate reacted?
It was already calculated: 240 g.
 

Answers and Replies

  • #2
Borek
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Could be the final question (about the mass of carbonate that reacted) asks how much carbonate reacted to give 34L of CO2 assuming 100% yield. It is not clear to me.
 
  • #3
205
3
Could be the final question (about the mass of carbonate that reacted) asks how much carbonate reacted to give 34L of CO2 assuming 100% yield. It is not clear to me.

Thank you for the response.
If the calcium carbonate is the limiting reagent, then the mass that reacted is simply the quantity given in the enunciation (considering the purity), right? Unfortunately, we would have to wait until Friday to ask exactly what the teacher meant.
It is strange that CaCO3 was the limiting reagent. If it was the one in excess, then the last question would appear more appropriate, because then we would have to find how many of the CaCO3 was actually used...
Anyway, is the yield of 45,6% right?
 
  • #4
205
3
We confirmed it with the teacher: 240 g is the right answer.
 

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