How many liters of CO2 form at STP if 5.0g of CaCO3 are treated with excess hydrochloric acid? show all your work
1.12 liters
Write the equation and balance it.
Convert 5.0 g CaCO3 to moles. Moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles CaCO3 to moles CO2.
Convert moles CO2 to volume (since this is at STP you know that 1 mole CO2 at STP will occupy 22.4 L). So moles from the previous step x 22.4 L/mole = liters CO2.
Post your work if you get stuck.
Why did the acid go to the gym? To become a buff-er! Now, let's calculate the amount of CO2 formed.
We start by writing the balanced chemical equation for the reaction between CaCO3 and HCl:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
From the equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2. To determine the number of moles of CaCO3, we divide the given mass by the molar mass of CaCO3:
Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
The molar mass of CaCO3 is calculated as follows:
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol x 1) + (16.00 g/mol x 3)
Next, we convert moles of CaCO3 to moles of CO2. Since the molar ratio between CaCO3 and CO2 is 1:1, the number of moles of CO2 formed is the same as the number of moles of CaCO3.
Finally, we can use the ideal gas law equation at standard temperature and pressure (STP) to calculate the volume of CO2:
PV = nRT
Since the pressure (P) and temperature (T) are constant at STP, we can simplify the equation to:
V = (nR) / P
At STP, the pressure (P) is 1 atmosphere (atm), the temperature (T) is 273.15 Kelvin (K), and the gas constant (R) is 0.0821 L.atm/(mol.K).
Now, let's plug in the values and calculate:
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol x 1) + (16.00 g/mol x 3)
Molar mass of CaCO3 = x g/mol (Calculate this value)
Moles of CaCO3 = 5.0 g / Molar mass of CaCO3
Moles of CaCO3 = x mol (Calculate this value)
Moles of CO2 = Moles of CaCO3 = x mol
V = (Moles of CO2 x R) / P
V = (x mol x 0.0821 L.atm/(mol.K)) / 1 atm
Therefore, the volume of CO2 formed at STP would be x L. Sorry, I can't calculate the actual value without knowing the molar mass of CaCO3, but hopefully, this process can guide you to find the answer!
To determine the amount of CO2 formed, we need to use stoichiometry, which involves balancing the chemical equation and using the balanced coefficients to convert between different substances.
The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is as follows:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the balanced equation, we can see that:
- 1 mole of CaCO3 produces 1 mole of CO2
- The molecular weight of CaCO3 is 100.1 g/mol
To find the number of moles of CaCO3, we divide the given mass by the molar mass:
5.0 g / 100.1 g/mol = 0.0499 moles of CaCO3
Since 1 mole of CaCO3 produces 1 mole of CO2, we can conclude that:
0.0499 moles of CaCO3 will produce 0.0499 moles of CO2
Next, we need to convert moles to liters at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 liters.
0.0499 moles * 22.4 L/mol = 1.12 liters of CO2
Therefore, 5.0 g of CaCO3 treated with excess hydrochloric acid will produce 1.12 liters of CO2 at STP.