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What mass of NaCrO2 can be obtained from the reaction of 7.40g Cr(OH)3 with 7.60g NaOH in the following reaction: CrOH3+NaOH==>NaCrO2+2H2O

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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

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