Answer:
17.4 grams butane, C4H10
Step-by-step explanation:
I'll assume the water, H2O, is produced from the butane, C4H10, by a combustion process using oxygen, O2. We need a balanced equation to start the calculation.
The basic combustion reaction is:
C4H10 + O2 = CO2 + H2O
The basic equation needs to be balanced. The same numbers and types of atoms must be equal on both sides. We can see that one atom of C4H10 contains 4 carbons and 10 hydrogens.
Carbon: The carbons wind up in the CO2 molecule. But each CO2 only has 1 carbon, so we need at least 4 CO2 molecules:
C4H10 + O2 = 4CO2 + H2O
Hydrogen: Hydrogen atoms are used to form the water. 10 hydrogens from 1 C4H10 will be used to produce 5 H2O molecules, since each requires 2 hydrogens.
The leads to:
C4H10 + O2 = 4CO2 + 5H2O
Now all the carbons and hydrogens are accounted for. But the oxygens are still not balanced. As written, we have 1 molecule of O2, for 2 oxygen atoms. But from the last step in our balancing work, we need 8 from the 4CO2) and 5 for the H2O, for a total of 13 oxygen atoms. Normally, this would be easy to add more oxygen molecules, but since they comes in pairs (O2), we can only add two at a time, so there will never be an odd number of O atoms from the reactant side.
To solve this, try adding a coefficient of 2 to the C4H10 and rebalancing the H2O and CO2:
2C4H10 + O2 = 4CO2 + 5H2O
We need to adjust both the CO2 and H2O to accommodate the added C and H atoms (a total of 8C's and 20 H's):
2C4H10 + O2 = 8CO2 + 10H2O
Now we see we need more O2, since there is a total of 16 plus 10, or 26 O atoms in the 8CO2 and 10H2O molecules. To provide 26 O atoms, we need a coefficient of 13 in front of the O2 on the reactant side:
2C4H10 + 13O2 = 8CO2 + 10H2O
Now the equation is balanced. It tells us we'll need 2 moles of C4H10 to produce 10 moles of H2O. That's a molar ratio of 5 moles water/1 mole butane. Or, (0.2 moles butane)/(mole water).
Now convert 27 grams water into moles water:
The molar mass of water is 18 grams/mole. 27 grams of H2O is therefore:
(27 grams)(18 grams/mole) = 1.5 moles H2O
We discovered earlier that we need (1/5) that number of moles of butane. Therefore, we need:
(1.5 moles H2O)*((0.2 moles butane)/(mole water)) = 0.3 moles butane to produce 1.5 moles (or 27 grams) of water.
Now convert 0.3 moles butane into grams butane by multiplying by butane's molar mass of 58.0 g/mole:
(0.3 moles)(58.0 g/mole) = 17.4 grams butane to produce 27 grams of water.
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Wheh! That's a lot of work for 27 grams of water. But the heat released will be fun.
