anonymous 5 years ago If a sample of Al2(SO4)3 is found to contain 5.57 · 10-3 mol of Al3+ ions how many moles of Al2(SO4)3 formula units are present? Express your answer in scientific notation.

1. anonymous

You know that one molecule of aluminium sulfate can produce two cations of Al3+, so$2 Al^{3+} \iff Al_2(SO_4)_3$which means one Al3+ is equivalent to 1/2 of one aluminium sulfate:$Al^{3+} \iff \frac{1}{2}Al_2(SO_4)_3$You have 5.55 x 10^-3 mol of Al3+, so you have$5.57 \times 10^{-3}mol Al^{3+} \iff \frac{1}{2}5.57 \times 10^{-3}molAl_2(SO_4)_3$that is, you have$2.785 \times 10^{-3}molAl_2(SO_4)_3$

2. anonymous

To keep to the number of significant figures, the number of moles of aluminium sulfate is $2.79 \times 10^{-3}mol$