Question

5g of H2 react with 24.5g of O2 to form H20. a) Which is the limiting reagent.

b) Calculate the amount H20 produced.

c) Which reactant remains uncreated.​

Answer 1

Explanation:

2Mg+O

2

→2MgO

The atomic masses of Mg and O are 24.3 g/mol and 16 g/mol respectively.

The molar mass of MgO=24.3+16=40.3g/mol

Number of moles is the ratio of mass to molar/atomic mass.

48.6 g of Mg corresponds to

24.3

48.6

=2 moles

64 g of O2 corresponds to

2×16

64

=2 moles

2 moles of Mg will react with 1 mole of O

2

to produce 2 moles of MgO.

Since 2 moles of O

2

are present, O

2

is excess reagent and Mg is limiting reagent.

2 moles of Mg will give 2 moles of MgO.

Mass of MgO =2×40.3=80.6g

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Answer 2

Answer:

a)O2

b)27.5

c)hydrogen excess