Question

If we want to form 6.25 grams of solid nickel how many grams of nickel (lll) oxide will we need to start with?

Answer 1

Answer:

8.77 g of Ni2O3.

Explanation:

The formation of solid nickel is given by the following reaction:

$2Ni_2O_3\rightarrow4Ni+3O_2.$

And we want to know how many grams of nickel (III) oxide (Ni2O3) we will require to form 6.25 g of Nickel (Ni). So, first, let's find the number of moles of 6.25 g of Ni using its molar mass which is 58.7 g/mol:

$6.25\text{ g Ni}\cdot\frac{1\text{ mol Ni}}{58.7\text{ g Ni}}=0.106\text{ moles Ni.}$

We can state a rule of three based on the chemical equation. You can see that 2 moles of Ni2O3 reacted produces 4 moles of Ni:

$\begin{gathered} 4\text{ moles Ni}\rightarrow2\text{ moles Ni}_2O_3 \\ 0.106\text{ moles Ni}\rightarrow?\text{ moles Ni}_2O_3 \end{gathered}$

The calculation would be:

$0.106\text{ moles Ni}\cdot\frac{2\text{ moles Ni}_2O_3}{4\text{ moles Ni}}=0.053\text{ moles Ni}_2O_3.$

The next step is to convert 0.053 moles of Ni2O3 to grams using the molar mass of Ni2O3 which is 165.4 g/mol:

$0.053\text{ moles Ni}_2O_3\cdot\frac{165.4\text{ g Ni}_2O_3}{1\text{ mol Ni}_2O_3}=8.7\text{7 g Ni}_2O_3.$

We will require 8.77 g of Ni2O3 to form 6.25 g of solid nickel.