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Acid precipitation dripping on Limestone produces Carbon Dioxide by the following reaction:
CaCo3(s) + 2H+(aq) ---> Ca(2+)(aq) + CO2(g) + H2O (l)
15.5mL of CO2 was produced at 25*C and 738.0 mmHg
How man moles of CO2 were produced?
How many milligrams of CaCO3 were consumed?

Respuesta :

1) PV=nRT
P=738.0 mmHg
V=15.5mL=0.0155 L
T=273+25=298 K
R=62.36 L*mmHg*K⁻¹mol⁻¹
n=PV/RT
n=(738.0 mmHg *0.0155 L)/(62.36 L*mmHg*K⁻¹mol⁻¹*298 K)= =0.000616=6.16*10⁻⁴ mol

2)  From the equation of the reaction 
        1 mol CaCO3 gives 1 mol CO2,
so 6.16*10⁻⁴ mol CaCO3 ---->  6.16*10⁻⁴ mol CO2

Molar mass CaCO3 =M(Ca)+M(C)+3*M(O)= 40.1+12.0+3*16.0 =100.1 g/mol
6.16*10⁻⁴ mol CaCO3 * 100.1 g/mol =617*10⁻⁴ g =0.0617 g = 61.7mg

Answer: The number of moles of carbon dioxide produced are [tex]6.16\times 10^{-4}mol[/tex] and the mass of calcium carbonate is 61.6 mg

Explanation:

To calculate the moles of carbon dioxide gas, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 738.0 mmHg  

V = Volume of the gas = 15.5 mL = 0.0155 L   (Conversion factor: 1 L = 1000 mL)

T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of carbon dioxide gas = ?

Putting values in above equation, we get:

[tex]738.0mmHg\times 0.0155L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{738\times 0.0155}{62.3637\times 298}=6.16\times 10^{-4}mol[/tex]

For the given chemical equation:

[tex]CaCO_3(s)+2H^+(aq.)\rightarrow Ca^{2+}(aq.)+CO_2(g)+H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of carbon dioxide gas is produced from 1 mole of calcium carbonate

So, [tex]6.16\times 10^{-4}mol[/tex] of carbon dioxide will be produced from = [tex]\frac{1}{1}\times 6.16\times 10^{-4}=6.16\times 10^{-4}mol[/tex] of calcium carbonate

  • To calculate the mass for given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = [tex]6.16\times 10^{-4}mol[/tex]

Putting values in above equation:

[tex]6.16\times 10^{-4}mol=\frac{\text{Mass of calcium carbonate}}{100g/mol}\\\\\text{Mass of calcium carbonate}=(6.16\times 10^{-4}mol\times 100g/mol)=6.16\times 10^{-2}g[/tex]

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So, [tex]6.16\times 10^{-2}g\times \frac{1000mg}{1g}=61.6mg[/tex]

Hence, the number of moles of carbon dioxide produced are [tex]6.16\times 10^{-4}mol[/tex] and the mass of calcium carbonate is 61.6 mg

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