To find the mass of the reactant (zinc carbonate), we first need to write and balance the chemical equation for the reaction:
ZnCO3 → ZnO + CO2
From the balanced equation, we can see that 1 mole of zinc carbonate (ZnCO3) produces 1 mole of zinc oxide (ZnO) and 1 mole of carbon dioxide (CO2).
1 mole of ZnCO3 has a molar mass of:
Zinc (Zn): 65.38 g/mol
Carbon (C): 12.01 g/mol
Oxygen (O): 16.00 g/mol
Molar mass of ZnCO3 = 65.38 + 12.01 + (3 x 16.00) = 125.39 g/mol
Now, we can calculate the number of moles of ZnCO3 by using the masses given for ZnO and CO2:
Number of moles of ZnO = 47.5 g / molar mass of ZnO (ZnO: 81.38 g/mol)
Number of moles of CO2 = 32.1 g / molar mass of CO2 (CO2: 44.01 g/mol)
Since 1 mole of ZnCO3 produces 1 mole of ZnO and 1 mole of CO2, the number of moles of ZnCO3 used in this reaction would be equal to the number of moles of ZnO or CO2. So, the number of moles of ZnCO3 can be taken as the number of moles of ZnO or CO2, which is:
Number of moles of ZnCO3 = 47.5 g / 81.38 g/mol = 0.584 moles
Number of moles of ZnCO3 = 32.1 g / 44.01 g/mol = 0.729 moles
Since these values are approximately equal, we can take either moles of ZnO or CO2 as the moles of ZnCO3. So, the mass of ZnCO3 can be calculated as follows:
Mass of ZnCO3 = moles of ZnCO3 x molar mass of ZnCO3
Mass of ZnCO3 = 0.584 moles x 125.39 g/mol
Mass of ZnCO3 = 73.20 grams
Therefore, the mass of the reactant (zinc carbonate) is approximately 73.20 grams.
If 47.5 grams of zinc (II) oxide and 32.1 grams of CO2 are produced in the reaction, the mass of the reactant (zinc carbonate) is how many grams?
1 answer