Question

How many grams of aluminum dichromate are formed from the double displacement reaction between 107.5g of aluminum nitrate and excess sodium dichromate? The products are aluminum dichromate and sodium nitrate (show work)

Answer 1

The reaction can be shown as following:

2Al(NO₃)₃ + 3Na₂Cr₂O₇ -->  Al₂(Cr₂O₇)₃ + 6NaNO₃

The amounts of aluminum dichromate produced in the reaction are the same as the amounts of aluminum nitrate used during the reaction. As a result:

m(Al₂(Cr₂O₇)₃) = m(Al(NO3)₃) × Mr(Al₂(Cr₂O₇)₃) / [2 × Mr(Al(NO₃)₃)]= 107.5 g × 702 g/mol / [2 × 213 g/mol] = 177.15 g

Answer: 177.15 g