## Chemistry 10th Edition

(a) $Al_2O_3 + 3H_2SO_4 -- \gt Al_2(SO_4)_3 + 3H_2O$ (b) $7.8mol(H_2SO_4)$. (c) $2.6 mol(Al_2(SO_4)_3)$
(a) Write the reaction, and balance it: $Al_2O_3 + H_2SO_4 -- \gt Al_2(SO_4)_3 + H_2O$ 1. Begin by balancing the elements that appear only once in each side: - Balance the $S$: The products side has a total of 3 sulfurs in $Al_2(SO_4)_3$. So, we should put that number as the coefficient of $H_2SO_4$. $Al_2O_3 + 3H_2SO_4 -- \gt Al_2(SO_4)_3 + H_2O$ - Balance the $H$: The reactants side has a total of 6 hydrogens in $3H_2SO_4$, so, to balance that, we should put a $\frac{6}{2}$ (which is 3) as the coefficient of $H_2O$ *We divide it by 2, because each water molecule has 2 hydrogens. $Al_2O_3 + 3H_2SO_4 -- \gt Al_2(SO_4)_3 + 3H_2O$ - The equation is balanced. ------------------- Use the coefficients as conversion factors: (b) According to the balanced equation, for each mole of $Al_2O_3$, it is necessary to have 3 moles of $H_2SO_4$: $2.6 mol(Al_2O_3) \times \frac{3mol(H_2SO_4)}{1mol(Al_2O_3)} = 7.8mol(H_2SO_4)$. (c) (b) According to the balanced equation, for each mole of $Al_2O_3$, we produce 1 mole of $Al_2(SO_4)_3$: $2.6mol(Al_2O_3) \times \frac{1mol(Al_2(SO_4)_3)}{1mol(Al_2O_3)}= 2.6 mol(Al_2(SO_4)_3)$