Question

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?

Answer 1

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 2

Answer: The number of moles of carbon dioxide produced are
`6.16* 10^(-4)mol` and the mass of calcium carbonate is 61.6 mg

Step-by-step 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* 0.0155L=n* 62.3637\text{ L.mmHg }mol^(-1)K^(-1)* 298K\\\\n=(738* 0.0155)/(62.3637* 298)=6.16* 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* 10^(-4)mol` of carbon dioxide will be produced from =
`(1)/(1)* 6.16* 10^(-4)=6.16* 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* 10^(-4)mol`

Putting values in above equation:

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

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,
`6.16* 10^(-2)g* (1000mg)/(1g)=61.6mg`

Hence, the number of moles of carbon dioxide produced are
`6.16* 10^(-4)mol` and the mass of calcium carbonate is 61.6 mg