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3 years ago

Calculate the mass of hydrogen in 500 g of Al(C2H3O2)3

2 answers:

4 0

22.2 grams For every molecule of Al(C2H3O2)3, we have 9 atoms of hydrogen. So let's first determine how many moles of Al(C2H3O2)3 we have. So start by looking up the atomic weights of the elements involved. Atomic weight aluminium = 26.981539 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass Al(C2H3O2)3 = 26.981539 + (12.0107 * 2 + 1.00794 * 3 + 15.999 * 2)*3 = 204.111199 Moles Al(C2H3O2)3 = 500 g / 204.111199 g/mol = 2.449645107 mol Moles H = 9 * 2.449645107 mol = 22.04680597 mol Mass H = 22.04680597 mol * 1.00794 g/mol = 22.22185761 g Rounding to 3 significant figures gives 22.2 grams.

Zigmanuir [339]3 years ago

3 0

Answer is: the mass of hydrogen is 22,05 grams.
m(Al(C₂H₃O₂)₃) = 500 g.
M(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.
n(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.

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Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

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$E^{o}_{AgCl/Ag}$ = 0.222 V

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