Calculate the equivalent mass of sulphuric acid(ii) The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AI= 27 u Atomic mass of O = 16 u) 2Al + Fe_(2)O_(3)toAI_(2)O_(3) + 2Fe, If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide. (a) Calculate the mass of AI_(2) O_(3) formed (b) How much of the excess reagent is left at the end of the reaction?

Calculate the equivalent mass of sulphuric acid(ii) The reaction betwe

1.	Calculate the equivalent mass of sulphuric acid(ii) The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AI= 27 u Atomic mass of O = 16 u) 2Al + Fe_(2)O_(3)toAI_(2)O_(3) + 2Fe, If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide. (a) Calculate the mass of AI_(2) O_(3) formed (b) How much of the excess reagent is left at the end of the reaction?
Answer» Solution :Equivalent mass of sulphuric acid:Sulphuric acid = `H_(2)SO_(4)` MOLAR mass of Sulphuric acid = `2 + 32 + 64 = 96` Basicity of Sulphuric acid = 2 Equivalent mass of acid = `("Molar mass of an acid")/("Basicity")=(96)/(2)=49geq^(-1)` (a) `underset(54g)(2A1)+underset(160g)(Fe_(2)O_(3))tounderset(102g)(A1_(2)O_(3))+underset(112g)(2Fe)` As per balanced equation 54 g Al is required for 112 g of iron and 102 g of `A1_(2)O_(3)54g` of Al gives 102 g of `Al_(2)O_(3)` `:.` 324 g of Al will give `(102)/(54)xx 324 = 612` g of `Al_(2)O_(3)` (b) 54 g of Al requires 160 g of `Fe_(2)O_(3)` for welding reaction therefore 324 g of Al will require `(160)/(54)xx 324` = 960 g of `Fe_(2)O_(3)` therefore Excess`Fe_(2)O_(3)`- Unreacted `Fe_(2))O_(3),= 1120 - 960 = 160 g` 160 g of excess reagent is left at the end of the reaction.