Question

Question #f3760

Answer 1

A good tool to have in your posession for such problems is the mole-to-mole ratio between the species involved in a reaction. Mole ratios can be set up between different reactants, different products, or between reactants and products.

Let's work an example to better illustrate this:

How many grams of methane must be burned in order to produce 15.6 g of `H_2O`?

`CH_4 + 2O_2 -> CO_2 + 2H_2O`

Notice that we have a `1:2` mole ratio between methane (`CH_4`) and water (`H_2O`); this means that 1 mole of methane will produce 2 moles of water. This can be set up

`15.6` `"g"` `H_2O * ("1 mole")/("18.0 g") * ("1 mole" CH_4)/("2 moles" H_2O) = 0.433` `"moles"` `CH_4`

This amounts to a mass of methane equal to

`0.433` `"moles" * ("16.0 g")/("1 mole") = 6.93` `"g"` `H_2O`

However, this example assumes a 100% yield for this reaction, something that does not happen in actual reactions.

Let's take the same example, but add that the reaction has a 83.3% yield. 15.6 g of water are now produced at a smaller yield, which means that more methane must react this time.

`"% yield" = ("actual yield")/("theoretical yield")`

We know that a 83.3% yield produces 15.6 g of water, so this is our practical yield. We can determine how much would be produced for a 100% yield:

`"% yield" = ("15.6 g")/(m_("theoretical")) * 100% =>`

`m_("theoretical") = (100 * 15.6)/(83.3) = 18.7` `"g"`

This means that the mass of methane that actually reacts

`18.7` `"g"` `H_2O * ("1 mole"H_2O)/("18.0 g") * ("1 mole"H_2O)/("2 moles"CH_4) * ("16.0 g"CH_4)/("1 mole"CH_4)`

equals `8.31` `"g of"` `CH_4`

Indeed, more methane is needed this time to produce the same amount of water because of the difference in percent yield.

Notice that the mole ratio between methane and water was used in this cas as well. So, if a reaction's percent yield is not specified, assume it to be 100%. If it is specified, use it to determine the actual quantity that reacts (or is produced).