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6CuNO3 + Al2(SO4)3 → 3Cu2SO4 + 2Al(NO3)3
Molar mass of CuNO3 125.56 g/mol
Molar mass of Al(NO3)3 213.01 g/mol

How many grams of copper (I) nitrate (CuNO3) are required to produce 44.0 grams of aluminum nitrate (Al(NO3)3)?

Sagot :

keziakh004

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.

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