In the following reaction, how many liters of carbon dioxide, CO2, measured at STP, would be produced from the decomposition of 143 g of magnesium carbonate, MgCO3?
MgCO3---MgO+CO2
how many moles of MgCO3 is in 143 grams?
1.70
then what is 1.70*22.4 liters
To determine the number of liters of carbon dioxide (CO2) produced from the decomposition of 143 g of magnesium carbonate (MgCO3) at STP (Standard Temperature and Pressure), we need to follow a few steps.
Step 1: Convert the mass of MgCO3 to moles.
To do this, we use the molar mass of MgCO3, which is the sum of the atomic masses of its constituent elements:
Mg: 24.31 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Therefore, the molar mass of MgCO3 is 24.31 + 12.01 + (16.00 x 3) = 84.31 g/mol.
To convert the mass of MgCO3 to moles, we divide the given mass by the molar mass:
Number of moles = Mass of MgCO3 / Molar mass of MgCO3
Number of moles = 143 g / 84.31 g/mol
Step 2: Determine the ratio of moles between MgCO3 and CO2.
From the balanced chemical equation, we can see that for every 1 mole of MgCO3, 1 mole of CO2 is produced. Therefore, the ratio of moles between MgCO3 and CO2 is 1:1.
Step 3: Convert the moles of CO2 to liters at STP.
At STP, 1 mole of any gas occupies a volume of 22.4 liters. Therefore, the number of liters of CO2 produced will be the same as the number of moles of CO2.
Now, let's calculate the answer.
Number of moles of MgCO3 = 143 g / 84.31 g/mol (divide the given mass by the molar mass)
Number of moles of CO2 = Number of moles of MgCO3 (since the ratio is 1:1)
Number of liters of CO2 = Number of moles of CO2 (since 1 mole of gas is equivalent to 22.4 liters at STP)
By following these steps, you can determine the number of liters of carbon dioxide produced from the given decomposition reaction.