<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
4.22 grams.
1. First find out how much AgNO3 weighs with one mole (107.87 g Ag + 14.007 g N + 48 g O = 169.89 grams)
2. Find the percent of Ag you have. So, (107.87 g/mol Ag)/(169.89 g/mol AgNO3)= 0.63 * 100 = 63%.
3. If you have 6.7 grams total, you know 63% of it is going to be silver, so just multiply 6.7 grams by .63 and you get 4.22 g Ag
To change only one variable which is very important than to test the experiment to match the hypothesis again, I think. It’s been a while since I was on that lesson♀️
