Question

Question #cffa0

Answer 1

When you are dealing with gases, it is necessary to have values for pressure and temperature. SInce no values are provided, I'll assume you are at STP - 1 atm and 273.15 K.

What basically takes place is the combustion of sucrose, or table sugar.

`C_12H_22O_11 + 12O_2 -> 11H_2O + 12CO_2`

The first thing you must look for is mole ratios, that will give you an idea on how to proceed once you determine how many moles of sucrose you have.

Notice that every 1 mole of sucrose needs 12 moles of oxygen for the reaction to take place - you have a `"1:12"` mole ratio between sucrose and oxygen. This tells you that whatever number of moles of sucrose you have, you'll need 12 times more moles of oxygen.

Determine the moles of sucrose you have by using its molar mass - `"342.3 g/mol"`

`"2.48 g" * ("1 mole")/("342.3 g") = "0.00725 moles sucrose"`

We've established that you need 12 times more moles of oxygen, so

`"0.00725 moles sucrose" * ("12 moles oxygen")/("1 mole sucrose") = "0.0870 moles"`

Now that you know how mnay moles of oxygen you have, you can use the fact that 1 mole of any ideal gas occupies exactly `"22.4 L"` at STP - this is known as the molar volume of a gas.

If 1 mole occupies `"22.4 L"`, and you only have `"0.0870 moles"`, the volume of oxygen you'll need will be obviously smaller than `"22.4 L"`

`"0.0870 moles" * ("22.4 L")/("1 mole") = "1.95 L oxygen"` `->` at STP.

If you're not at STP, use the ideal gas law equation to solve for `"V"`:

`PV = nRT => V = (nRT)/P` - then just plug in whatever values you have for pressure and temperature.

Answer 2

2.09 litres of `O_2` are required.

The `M_r` of the carbohydrate from which the "gummy bear" is composed `=[(12xx12)+(22xx1)+(11xx16)]=342`

I will assume we are at NTP where the volume of 1 mole of gas = 24 litres.

From the equation we can see that:

1 mole of gummy bear needs 12 moles `O_2`

1 mole needs `12xx24=288` litres `O_2`

So 342g needs 288 litres `O_2`

So 1g needs `(288)/(342)` litres `O_2`

So 2.68g needs `(288)/(342)xx2.48=2.09` litres