Question

Acid precipitation dripping on Limestone produces Carbon Dioxide by the following reaction:

Answer 1

Answer: The number of moles of carbon dioxide produced are $6.16\times 10^{-4}mol$ 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:

$PV=nRT$

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 = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of carbon dioxide gas = ?

Putting values in above equation, we get:

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

For the given chemical equation:

$CaCO_3(s)+2H^+(aq.)\rightarrow Ca^{2+}(aq.)+CO_2(g)+H_2O(l)$

By Stoichiometry of the reaction:

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

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

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = $6.16\times 10^{-4}mol$

Putting values in above equation:

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

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So, $6.16\times 10^{-2}g\times \frac{1000mg}{1g}=61.6mg$

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

Answer 2

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