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adoni [48]
3 years ago
5

35.10 g of aluminum hydroxide is allowed to react with 53.94 g of sulfuric acid as follows:2Al(OH)3(s) + 3H2SO4(aq) -------->

Al2(SO4)3(aq) + 6H2O(l)What is the maximum mass of aluminum sulfate which could be produced in the reaction?
Chemistry
1 answer:
ikadub [295]3 years ago
8 0

Answer:

62.586 gram

Explanation:

moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol

moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol

H2SO4 is the limiting reagent (reacts completely)

⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4

moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol

mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram

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Explanation:

The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:

When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.

For compounds, you add the atomic masses of the component atoms and you sum up.

You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P

Atomic mass of Mg = 24.3g/mol

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Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol

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Explanation:

Hello,

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In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:

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