What mass of NaCrO2 can be obtained from the reaction of 7.40 g Cr(OH)3 with 7.60 g NaOH in this reaction Cr(OH)3+ NaOH->NaCrO2+2H2O what is the limiting reactant

Answer 1

Cr(OH)₃ + NaOH ==> NaCrO₂ + 2H₂O

Moles Cr(OH)₃ present = 7.40 g x 1 mole/103 g = 0.0718 moles

Moles NaOH present = 7.60 g x 1 mol/40 g = 0.0543 moles

Limiting reactant = NaOH since mole ratio in balanced reaction is 1:1 and there is less NaOH than Cr(OH)₃

Theoretical yield of NaCrO₂ = 0.0543 moles NaOH x 1 mole NaCrO₂/mole NaOH x 107 g/mole = 5.81 g