At STP 12.78 L of CO2 is produced in 150.0 seconds. What is the rate of consumption of C3H7OH in g/min?
2 C3H7OH(l) + 9 O2(g) --> 6 CO2(g) + 8 H2O(g) + heat
At STP 12.78 L of CO2 is produced in 150.0 seconds.
What is the rate of consumption of C3H7OH in g/min?
- Lexi RLv 79 years agoBest Answer
Work out how many moles of CO2 are produced in 150.0 seconds
Convert this to moles CO2 produced per minute
Then determine how many moles of C3H7OH are needed to produce this many moles of CO2, this will be the number of moles of C3H7OH that are consumed per minute.
Convert moles C3H7OH / min to g/min.
At STP 1 mole of any ideal gas occupies a volume of 22.4 L
So moles CO2 produced = 12.78 L / 22.4 L/mol
= 0.5705 mol
So 0.5705 moles are produced in 150.0 sec
0.5705 moles every 2.5 minutes
= 0.2282 moles / min of CO2 are produced
The balanced equation shows that
6 moles CO2 are formed from 2 moles C3H7OH
So 1 mole CO2 forms from 2/6 moles C3H7OH
Thus 0.2282 moles CO2 forms from (2/6 x 0.2282) moles C3H7OH
= 0.07607 moles C3H7OH
Thus 0.07607 moles of C3H7OH are consumed every minute to produce 0.2282 moles CO2 every minute
mass = molar mass x moles
mass C3H7OH = 60.094 g/mol x 0.07607 mol
= 4.571 g
4.571 g/min are consumed