Question

III. (4 points) A sample of nickel ore, which has nickel as the only metal present, is treated with an excess of sulfuric acid (H2SO4) to form nickel(II) sulfate and molecular hydrogen. (a) Write a balanced equation for the reaction. (b) If 0.0764 g of H2 is obtained from 3.86 g of the nickel ore, calculate the percent nickel, by mass, of the nickel ore.

Answer 1

1. balanced equation for the reaction:

`Ni(s) +H_2SO_4(aq) \rightarrow NiSO_4(aq) +H_2(g)`

2. percentge of nickel=58.1%

Step-by-step explanation:

As given in the question that only nickel metal is present so hydrogen gas is produced by only nickel and there will be other substance too but only nickel can produced hydrogen as as reaction between metal and acid gives hydrogen gas.

1.

balanced equation for the reaction:

`Ni(s) +H_2SO_4(aq) \rightarrow NiSO_4(aq) +H_2(g)`

2.

From stoichiometry, we know

1 mole H_2 formed from 1 mole of Ni(s)

therefore 2 gram formed from 58.7 gram of Ni(s)

0.0764 gram formed from
`(58.7)/(2) * 0.0764=2.24 gram`

2.24 gram pure Ni(s) needs to produce 0.0764 gram of hydrogen

but this much hydrogen is produced from 3.86 gram of ore.

`percentage\, of\,nickel =(pure\, substance\, need)/(impure \,substance ) * 100`

`percentage\, of\,nickel =(2.24)/(3.86) * 100=58.1`

percentge of nickel=58.1%