The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g
1.25 g of CaCO₃ is required
A conversion factor is ALWAYS equal to 1 :)
Answer:
V = 34.55 L
Explanation:
Given that,
No of moles, n = 1.4
Temperature, T = 20°C = 20 + 273 = 293 K
Pressure, P = 0.974 atm
We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :
PV=nRT
R is gas constant, 
Finding for V,
So, the volume of the gas is 34.55 L.
Glucose is the starting molecule for glycolysis.
