Respuesta :
Answer:
[tex]\%\ mass\ of\ CaCO_3=93.37\ \%[/tex]
Explanation:
Given that:
Pressure = 791 mmHg
Temperature = 20.0°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = 293.15 K
Volume = 100 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.3637 L.mmHg/K.mol
Applying the equation as:
791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K
⇒n of [tex]CO_2[/tex] produced = 0.0493 moles
According to the reaction:-
[tex]CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2[/tex]
1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts
0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts
Moles of calcium carbonate reacted = 0.0493 moles
Molar mass of [tex]CaCO_3[/tex] = 100.0869 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.0493\ mol= \frac{Mass}{100.0869\ g/mol}[/tex]
[tex]Mass_{CaCO_3}=4.93\ g[/tex]
Impure sample mass = 5.28 g
Percent mass is percentage by the mass of the compound present in the sample.
[tex]\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100[/tex]
[tex]\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100[/tex]
[tex]\%\ mass\ of\ CaCO_3=93.37\ \%[/tex]
