2C12H4CI6+23O2+2H2O->24CO2+12HCI (d) how many grams of C12H4CI6 are reacted when 108.53 g of CO2 are produced? Note: use three significant digits in your answer. g C12H4CI6
To determine the amount of C12H4CI6 reacted, we need to use stoichiometry and the molar mass of C12H4CI6.
The balanced equation tells us that the ratio between C12H4CI6 and CO2 is 2:24 (or 1:12). This means that for every 12 moles of CO2 produced, 1 mole of C12H4CI6 is reacted.
We can calculate the moles of CO2 produced using its molar mass of 44.01 g/mol:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 108.53 g / 44.01 g/mol ≈ 2.464 moles
From the ratio, we can see that 12 moles of CO2 corresponds to 1 mole of C12H4CI6.
Thus, the moles of C12H4CI6 reacted can be calculated as:
moles of C12H4CI6 = moles of CO2 / 12
moles of C12H4CI6 = 2.464 moles / 12 ≈ 0.205 moles
Finally, we can convert moles of C12H4CI6 to grams using its molar mass of 376.64 g/mol:
grams of C12H4CI6 = moles of C12H4CI6 * molar mass of C12H4CI6
grams of C12H4CI6 = 0.205 moles * 376.64 g/mol ≈ 77.25 g
Therefore, approximately 77.25 grams of C12H4CI6 are reacted when 108.53 g of CO2 are produced.