How many grams of water are produced when 10.0g of Butane, C4H10(g) completely combust with 2.50 moles of Oxygen gas?
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To determine the number of grams of water produced when 10.0g of butane (C4H10) completely combusts with 2.50 moles of oxygen gas (O2), we need to follow a few steps.
1. Write the balanced equation for the combustion reaction:
C4H10 + O2 → CO2 + H2O
According to the balanced equation, for every mole of butane burned, we get one mole of water produced.
2. Calculate the number of moles of butane:
Given that the mass of butane is 10.0g and its molar mass is approximately 58.12 g/mol, we can use the formula:
moles = mass / molar mass
moles of butane = 10.0g / 58.12 g/mol
3. Determine the limiting reactant:
To find the limiting reactant, we compare the moles of butane used with the moles of oxygen available. From the balanced equation, we know that the reaction ratio is 1:13 (one mole of butane to 13 moles of oxygen). Butane is the limiting reactant if it has fewer moles than what is required in the reaction.
4. Calculate the moles of water produced:
Since one mole of butane reacts to produce one mole of water, the number of moles of water produced will be the same as the number of moles of butane used.
5. Convert moles to grams:
To convert the moles of water to grams, we need to use its molar mass. The molar mass of water (H2O) is approximately 18.02 g/mol.
grams of water = moles of water × molar mass of water
By following these steps, you should be able to calculate the grams of water produced.