Question

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

Answer 1

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

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