Question

How many grams of copper (I) nitrate (CuNO3) are required to produce 44.0 grams of aluminum nitrate (Al(NO3)3)?Use this equation to answer: 6CuNO3 + Al2(SO4)3 → 3Cu2SO4 + 2Al(NO3)3

Answer 1

77.8 grams

Explanation

The given balanced chemical equation is:

`6\text{CuNO}_3+Al_2(SO_4)_3\rightarrow2Cu_2SO_4+2Al(NO_3)_3`

From the balanced chemical equation for the reaction, 6 moles of copper (I) nitrate (CuNO₃) produced 2 moles of aluminum nitrate (Al(NO₃)₃).

That is, mole ratio of CuNO₃ to Al(NO₃)₃ is 6 : 2

Note that:

Molar mass of CuNO₃ = 125.55 g/mol

Molar mass of Al(NO₃)₃ = 212.996 g/mol

For CuNO₃

It implies that 1 mole of CuNO₃ = 125.55 g

Therefore, 6 moles of CuNO₃ will be = 6 x 125.55 g = 753.30 g

Also for Al(NO₃)₃

It implies that 1 mole of Al(NO₃)₃ = 212.996 g

So 2 moles of Al(NO₃)₃ = 2 x 212.996 g = 425.992 g

Now, we shall calculate the number of grams of copper (I) nitrate (CuNO₃) required to produce 44.0 grams of aluminum nitrate (Al(NO₃)₃) as follows:

Let the number of copper (I) nitrate (CuNO₃) required to be x

`\begin{gathered} 753.30g\text{ }CuNO_3=425.992g\text{ }Al(NO_3)_3 \\ xg\text{ }CuNO_3=44.0g\text{ }Al(NO_3)_3 \\ To\text{ get }xg\text{ }CuNO_3,\text{ cross multiply and divide both sides by }425.992g\text{ }Al(NO_3)_3 \\ xg\text{ }CuNO_3=\frac{753.30gCuNO_3*44.0g\text{ }Al(NO_3)_3}{425.992g\text{ }Al(NO_3)_3} \\ xg\text{ }CuNO_3=77.8071g \\ xg\text{ }CuNO_3\approx77.8g \end{gathered}`

Hence, 77.8 grams of copper (I) nitrate (CuNO₃) are required to produce 44.0 grams of aluminum nitrate (Al(NO₃)₃).