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## 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 37

^{o}C and a pression of 0.82 atm. What is the yield of this reaction and what mass of calcium carbonate reacted? (Remember that H

_{2}CO

_{3}, in water, decomposes as water and CO

_{2})

[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; 1CaCO

_{3}= 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 CaCO

_{3}.

Now, the molar relation between calcium carbonate and the gas, CO

_{2}, in order to find how many litres of the gas (y) are produced with 100% yield.

But first, to find the molar volume:

37

^{o}C = 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.