To solve the given problem, we need to determine the limiting reagent, calculate the mass of CaCl2 formed, find the number of moles of water produced, and determine the volume of CO2 produced. Let's solve each part step by step:
Given:
Mass of CaCO3 = 200 g
Purity of CaCO3 = 90% (90% pure CaCO3 means the remaining 10% is impurities)
Molar mass of CaCO3 = 100.09 g/mol
Molar mass of CaCl2 = 110.98 g/mol
Molar mass of H2O = 18.02 g/mol
Molar mass of CO2 = 44.01 g/mol
Temperature (T) = 27°C
Pressure (P) = 760 mmHg
Step A: Determining the limiting reagent
To find the limiting reagent, we need to compare the moles of CaCO3 and HCl. First, we calculate the moles of CaCO3:
Moles of CaCO3 = (Mass of CaCO3 * Purity of CaCO3) / Molar mass of CaCO3
= (200 g * 0.90) / 100.09 g/mol
≈ 1.798 mol
Since there is excess HCl, we don't need to calculate the moles of HCl.
Step B: Calculating the mass of CaCl2 formed
From the balanced chemical equation, we know that the molar ratio of CaCO3 to CaCl2 is 1:1. Therefore, the moles of CaCl2 formed will be equal to the moles of CaCO3:
Moles of CaCl2 = 1.798 mol
Mass of CaCl2 = Moles of CaCl2 * Molar mass of CaCl2
= 1.798 mol * 110.98 g/mol
≈ 199.34 g
The mass of CaCl2 formed is approximately 199.34 g.
Step C: Finding the moles of water produced
From the balanced chemical equation, we know that the molar ratio of CaCO3 to H2O is 1:1. Therefore, the moles of water produced will be equal to the moles of CaCO3:
Moles of H2O = 1.798 mol
Step D: Determining the volume of CO2 produced
To find the volume of CO2 produced, we can use the ideal gas law:
PV = nRT
Where:
P = Pressure (in atm) = 760 mmHg / 760 = 1 atm
V = Volume of CO2 (in liters)
n = Moles of CO2
R = Ideal gas constant = 0.0821 L·atm/(mol·K)
T = Temperature in Kelvin = 27°C + 273.15 = 300.15 K
Rearranging the equation, we have:
V = (nRT) / P
Moles of CO2 = Moles of CaCO3
V = (1.798 mol * 0.0821 L·atm/(mol·K) * 300.15 K) / 1 atm
≈ 44.94 L
The volume of CO2 produced is approximately 44.94 liters.
To summarize:
A. The limiting reagent is CaCO3.
B. The mass of CaCl2 formed is approximately 199.34 g.
C. The number of moles of water produced is 1.798 mol.
D. The volume of CO2 produced at 27°C and 760 mm