Aluminum will react with sulfuric acid in the following reaction: $2 A l (s) + 3 H_{2} S O_{4} (l) \to A l_{2} {(S O_{4})}_{3} (a q) + 3 H_{2} (g)$ $2Al(s) + 3H_2SO_4(l) -> Al_2(SO_4)_3(aq) + 3H_2(g)$ . How many moles of $H_{2} S O_{4}$ $H_2SO_4$ will react with 18 mol Al? How many moles of each product will be produced?

Question

Aluminum will react with sulfuric acid in the following reaction: $2 A l (s) + 3 H_{2} S O_{4} (l) \to A l_{2} {(S O_{4})}_{3} (a q) + 3 H_{2} (g)$ $2Al(s) + 3H_2SO_4(l) -> Al_2(SO_4)_3(aq) + 3H_2(g)$ . How many moles of $H_{2} S O_{4}$ $H_2SO_4$ will react with 18 mol Al? How many moles of each product will be produced?

1 Answer

Impact of this question

19854 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T09:52:16

$= 27 m o l H_{2} S O_{4}$ $=27molH_2SO_4$

Explanation:

With reference to the given balanced equation for the mole ratios, the needed number of moles for each remaining compound involved in the reaction using the mass of $A l$ $Al$ as basis for the molar conversions can be computed as;

$η H_{2} S O_{4} :$ $etaH_2SO_4:$
$=18cancel(molAl)xx(3molH_2SO_4)/(2cancel(molAl))$
$=27molH_2SO_4$

$etaAl_2(SO_4)_3:$
$=18cancel(molAl)xx(1molAl_2(SO_4)_3)/(2cancel(molAl))$
$=9molAl_2(SO_4)_3$

$etaH_2:$
$=18cancel(molAl)xx(3molH_2)/(2cancel(molAl))$
$=27molH_2$

Science

Math

Humanities

... and beyond

1 Answer

Explanation:

Related questions

Impact of this question