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nirvana33 [79]
2 years ago
8

What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol)?

Chemistry
1 answer:
NemiM [27]2 years ago
3 0

the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g

The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn

3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2

the formula is n= mass/M so, now substituting values

m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3

so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g

so mass of aluminum oxide obtained = 1.36g

To learn more about Mass:

brainly.com/question/19694949

#SPJ4

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Answer:

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Explanation:

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<u></u>

<u>1. Number of moles of H</u><u>₂</u>

  • number of moles = mass in grams / molar mass
  • molar mass of H₂ = 2.016g/mol
  • number of moles = 6.00grams / 2.016g/mol = 2.97619mol

<u></u>

<u>2. Number of moles of NH</u><u>₃</u>

<u></u>

i) Chemical equation:

  • 3H₂(g) + N₂(g) →  2NH₃(g)

ii) Mole ratio:

  • 3 mol H₂ : 2 mol NH₃

iii) Proportion:

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  • x = 4.4642857mol NH₃

<u>3. Volume of NH₃</u>

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A

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