Please help me !

Question

Consider the reaction:

ZnO(s) + H₂O(l) → Zn(OH)₂(aq)

Given an initial mass of 16.31 g ZnO, an excess of H₂O, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of Zn(OH)₂ produced by the reaction.

Answer 1

Step 1: balance the equiation Equation is already balanced

StepII: Find the molar mass :

M(ZnO) = 65.38+ 16.00 = 81.38 g

M(Zn(OH)₂) = 65.38 + 16 x2 + 1.008 x 2 = 99.38 g

Flow : gm of ZnO -> moles of ZnO ->moles of mole Zn(OH)₂->gm of Zn(OH)₂

1 mole ZnO 1 mole Zn(OH)₂ 99.38 g Zn(OH)₂

16.31 g ZnO x------------------ x ----------------------x----------------------- = 19.92 g of Zn(OH)₂

81.38 g ZnO 1 mole ZnO 1 mole Zn(OH)₂