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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Answer:
very reactive non metals
Explanation:
keep working on her class
Using the rule Q=mcdeltaT
Q=0.3850j/g*22.8*875
Q=7505.19 J
Answer: A. 5.41
Explanation:
To calculate the moles :
According to stoichiometry :
3 moles of 
require = 2 moles of 
Thus 0.0261 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 3 moles of 
Thus 0.0261 moles of give =
of
Mass of 
Thus 5.41 g of solid lead will be produced from the given masses of both reactants.
