The balanced equation is shown below.
2 HCl(aq) +CaCO3(s)=CaCl2(aq) +H2O(l) +CO2(g)
2.00 dm3 of a 3.00 M hydrochloric acid solution is added to the kettle (this is an excess).
During the reaction 1.20 dm3 of CO2 is produced (assume 250C and 1 atmosphere)
How many moles of CO2 are produced?
If you only want the mols, that's PV = nRT and solve for n.
P = 1 atm
V = 1.2 L
R = 0.0206
T = 273 + 250 = ?K
ok I checked that the answer is 0.05 but i have no idea how it's done...
Anyone can show the steps?
You can't plug and chug?
PV = nRT
The problem says you have 1.2 dm^23 (which is 1L) CO2 at the conditions listed.
P = 1 atm
V = 1.2 L
n = ?
R = 0.08206 L*atm/mol*K
T = 273 + 250 = 523 K
(1)(1.2) = n*0.08206*523
n = (1*1.2)/(0.08206*523)
n = 0.02796 mols which I would round to 0.028 to two significant figures. Your 0.05 number is not right (or you copied the problem wrong or looked up the wrong answer).
To find the number of moles of CO2 produced, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's convert the given volume of CO2 from dm3 to liters since the ideal gas law equation requires volume in liters.
Given:
Volume of CO2 = 1.20 dm3
Since 1 dm3 is equal to 1 liter, the volume of CO2 can be directly converted to liters:
Volume of CO2 = 1.20 liters
Now, we need to find the number of moles of CO2. Rearranging the ideal gas law equation, we get:
n = PV / RT
We are given:
Pressure (P) = 1 atmosphere
Volume (V) = 1.20 liters
Temperature (T) = 25°C (250C is equivalent to 25°C)
Gas constant (R) = 0.0821 L·atm/(mol·K)
Let's convert the temperature from Celsius to Kelvin:
T (Kelvin) = T (Celsius) + 273.15
T (Kelvin) = 25°C + 273.15
T (Kelvin) = 298.15 K
Now we can calculate the number of moles of CO2:
n = (1.20 * 1) / (0.0821 * 298.15)
Using a calculator, we evaluate the expression:
n ≈ 0.0488 moles
Therefore, approximately 0.0488 moles of CO2 are produced.