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Explanation:
7.60 g of NaOH = 7.6/40 = 0.19 moles.
Reaction is 1:1 so NaOH in excess and Cr(OH)3 is limiting reactant.
You will get maximum of 0.0718 moles of NaCrO2 which is 0.0718*107 = 7.69 g