Chemistry, 03.08.2019 23:30 laureleiolson

Astudent performs a titration analysis for the following balanced reaction: 3 h2so4 + 2 al(oh)3 → al2(so4)3 + 6 h2o if a 200.0g sample of al(oh)3 is used to react with excess h2so4, how many grams of aluminum sulfate are produced? a 157g b 630g c 1750g d 438g

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The correct answer is 438g according to the following workings.

mole ratio of Aluminium hydroxide to aluminium sulphate= 2:1
number of moles in 200g of aluminium hydroxide = 200÷78
=2.564 moles
number of moles of Al2(SO4) produced by 200g of Al(OH)3=1.282
mass of Al2(SO4) produced = 342×1.282
= 438g

Answer from: rockinrachel9099

D. 438g
Looking at the balanced reaction, for every 2 moles of Al(OH)3 consumed, 1 mole of Al2(SO4)3 is produced. So let's start by calculating the molar mass of the reactant and product.
Atomic weight aluminum = 26.981539
Atomic weight sulfur = 32.065
Atomic weight oxygen = 15.999
Atomic weight hydrogen = 1.00794

Molar mass Al(OH)3 = 26.981539 + (15.999 + 1.00794)*3 = 78.002359 g/mol
Molar mass Al2(SO4)3 = 26.981539 * 2 + (32.065 + 15.999*4)*3 = 342.146078 g/mol

Moles Al(OH)3 = 200.0 g / 78.002359 g/mol = 2.564025019 mol
Moles Al2(SO4)3 = 2.564025019 mol / 2 = 1.282012509 mol

Grams Al2(SO4)3 = 1.282012509 mol * 342.146078 g/mol = 438.635552 g
And the closest answer is "D. 438g"

Answer from: Quest

i believe the answer is c. "baking" is the gerund, while "baking blueberry muffins" would be the gerund phrase. hope this ! : )

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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Astudent performs a titration analysis for the following balanced reaction: 3 h2so4 + 2 al(oh)3 → a...

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