Answer:
77.81 g of CuNO₃
Explanation:
6 CuNO₃ + Al₂(SO₄)₃ ⇒ 3 Cu₂SO₄ + 2 Al(NO₃)₃
This is your chemical equation. We will need this information for converting from Al(NO₃)₃ to CuNO₃.
First, convert grams of Al(NO₃)₃ into moles using the molar mass.
[tex]44.0g*\frac{mol}{213.01g} = 0.2065mol[/tex]
Next, convert moles of Al(NO₃)₃ to moles of CuNO₃. You can do this by using the stoichiometry (the numbers in front of the compounds). For every 6 moles of CuNO₃, you will get 2 moles of Al(NO₃)₃, causing the ratio of CuNO₃ to Al(NO₃)₃ to be 6:2.
[tex]0.2065mol*\frac{6}{2} =0.6197mol[/tex]
Now that you moles of CuNO₃, convert this to grams using its molar mass.
[tex]0.6197 mol * \frac{125.56g}{mol} =77.81 g[/tex]
You will need 77.81 grams of CuNO₃ to produce 44.0 grams of Al(NO₃)₃.
