# 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?

### 1 Answer

- Lexi RLv 71 decade agoFavorite 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

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