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Topic: Percentage yield: 2Al(OH)3 + 3H2SO4 goes to Al2SO3 + 6H20  (Read 5646 times)

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aimz_186

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Percentage yield: 2Al(OH)3 + 3H2SO4 goes to Al2SO3 + 6H20
« on: February 05, 2006, 08:01:57 PM »
HEy guys,
having trouble with this....

2Al(OH)3 + 3H2SO4 goes to Al2SO3 + 6H20

If 15.0g of Al(OH)2 and 20.0g of H2SO4 are mixed together, the number of grams of Al(OH)3 left unreated at the end of the reaction is;

a)0
b)4.4g
c)10.6g
d)11.7g

So far i have only worked out the moles of Al(OH)3 is 0.1923
and the moles of H2SO4 is .20389

from there im stuck... thanks in advance for any help
« Last Edit: February 06, 2006, 12:58:59 PM by Mitch »

kkrizka

• Guest
Re:Percentage yield??
« Reply #1 on: February 05, 2006, 09:13:41 PM »
Use stoichiometry. For every 2 moles of Aluminum Hydroxide, 3 moles of Sulfuric acid reacts. You can use dimensional analysis to find out the amount of Aluminum Hydroxide reacted:
.20389 mol H2SO4 x 2 mol Al(OH)3
-------------
3 mol H2SO4

Then just substract this amount from the initial amount of Aluminum Hydroxide and use molar mass to convert it back to grams.

AWK

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Re:Percentage yield??
« Reply #2 on: February 06, 2006, 04:37:11 AM »
Al2SO3  should be Al2(SO4)3
You can replace moles by masses of reactants. Since there is indicated which is a limiting reactant, you cen lay down the following equation
x/156 = 20/294 (note x is a mass of reacted Al(OH)3 )
AWK