For the following reaction:
Al2(SO4)3 + 6NaOH = 2Al(OH)3 + 3Na2SO4
How many moles of Na2SO4 can be made with 3.2 moles of NaOH and excess Al2(SO4)3?
To determine how many moles of Na2SO4 can be made, we need to establish a mole-to-mole ratio between NaOH and Na2SO4 based on the balanced chemical equation.
Looking at the equation:
Al2(SO4)3 + 6NaOH = 2Al(OH)3 + 3Na2SO4
We can see that 6 moles of NaOH react with 3 moles of Na2SO4.
To find out how many moles of Na2SO4 can be made from 3.2 moles of NaOH, we can set up a proportion using the mole-to-mole ratio:
(6 moles NaOH / 3 moles Na2SO4) = (3.2 moles NaOH / x moles Na2SO4)
Cross-multiplying, we get:
6 * x = 3 * 3.2
6x = 9.6
Dividing both sides by 6, we find:
x = 1.6
Therefore, with 3.2 moles of NaOH, you can produce 1.6 moles of Na2SO4 when Al2(SO4)3 is in excess.