Question

the volume of h2 at stp required to completely reduce 160 grams of fe2O3​

Answer 1

Answer:

Fe2O3+3H2→2Fe+3H2O

The molar mass of  Fe2O3 is 2(56)+3(16)=160 g/mol.

160 g of  Fe2O3 corresponds to 160g/mol160g=1mol

Complete reduction of 1 mole of  Fe2O3 will require 3 moles of H2.

3 moles of H2 will occupy a volume of 3×22.4 litres at STP.

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Answer 2

Answer:

Explanation:

By poac of o

3 moles fe2o3 = mole h2o

By poac of h

2 moles h2 = 2 moles h2o

3moles fe2o3 = moles h2

3=moles h2