If 18.5 g of NaOH is added to 0.800 L of 1.00 M Zn(NO₃)₂, what mass in grams of Zn(OH)₂ will be formed in the following precipitation reaction? 2 NaOH(aq) + Zn(NO₃)₂(aq) → Zn(OH)₂ (s) + 2 NaNO₃ (aq)