calculate the amount of 95% marble required to obtain 11.2Litres OF CARBON DIOXIDE
You should calculate fully and try to be a little more understanble
To calculate the amount of 95% marble required to obtain 11.2 liters of carbon dioxide (CO2), we need to know the balanced chemical equation for the reaction that produces carbon dioxide from marble. Assuming we are referring to the thermal decomposition of calcium carbonate (CaCO3), the balanced equation is:
CaCO3 -> CaO + CO2
From this equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2.
Step 1: Calculate the moles of CO2 required.
To find the moles of CO2, we can use the ideal gas law:
PV = nRT
Where:
P = pressure of CO2
V = volume of CO2
n = moles of CO2
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature
Assuming the temperature is constant and the pressure is atmospheric (1 atm), we can rearrange the equation:
n = PV / RT
For this calculation, we have:
V = 11.2 L
P = 1 atm
R = 0.0821 L·atm/mol·K
T = constant
n = (11.2 L * 1 atm) / (0.0821 L·atm/mol·K * T)
Please provide the temperature (in Kelvin) at which this reaction is occurring, so we can calculate the moles of CO2 more accurately.
Step 2: Convert moles of CO2 to moles of CaCO3.
Since the balanced equation shows a 1:1 molar ratio between CaCO3 and CO2, the number of moles of CaCO3 required to produce the calculated amount of CO2 will be the same.
Step 3: Calculate the mass of 95% marble required.
To find the mass of 95% marble required, we need to multiply the moles of CaCO3 by the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol.
Mass = moles of CaCO3 * molar mass of CaCO3
I hope this helps you understand how to calculate the amount of 95% marble required to obtain a given volume of carbon dioxide! If you provide the temperature, I can assist you further in completing the calculation.
CaCO3→CaO+CO2CaCO3→CaO+CO2
Given volume of CO2" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.2px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; CO2CO2 = 11.2 litres
Molar volume at STP = 22.4 litres
1 mole CO2 in the reaction = 1 mole CaCO3No. of moles of CO2 = frac11.222.4=0.5moles" role="presentation" style="box-sizing: border-box; display: table-cell !important; line-height: 0; text-indent: 0px; text-align: center; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.36px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 20.935em; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; width: 10000em; position: relative;">1 mole CO2 in the reaction = 1 mole CaCO3No. of moles of CO2 = frac11.222.4=0.5moles1 mole CO2 in the reaction = 1 mole CaCO3No. of moles of CO2 = frac11.222.4=0.5moles
1 mole  rmCO2" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.2px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;"> rmCO2 rmCO2 in the reaction = 1 mole CaCO_3
Therefore, 0.5 mole CO2 means 0.5 mol CaCO3 has reacted in the above reaction.
No. of grams of CaCO3=no. of moles of CaCO3×molar mass of CaCO3=0.5×100=50g" role="presentation" style="box-sizing: border-box; display: table-cell !important; line-height: 0; text-indent: 0px; text-align: center; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.36px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 31.886em; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; width: 10000em; position: relative;">No. of grams of CaCO3=no. of moles of CaCO3×molar mass of CaCO3=0.5×100=50gNo. of grams of CaCO3=no. of moles of CaCO3×molar mass of CaCO3=0.5×100=50g
Since the marble is only 95% pure,
Amount of marbleCaCO3=500.95=52.63g" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.2px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">CaCO3=500.95=52.63g
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
11.2 L CO2 (atSTP) = 11.2/22.4 = 0.5 mol.
1 mole CO2 in the reaction = 1 mole CaCO3 initially; therefore, 0.5 mole CO2 means you started with 0.5 mol CaCO3.
grams = moles x molar mass = ??
Since the marble is only 95% pure, you must start with a little more marble or
??g CaCO3 calculated/0.95 = xx.
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html