Please explain steps and why
First, we need to write a balanced equation for the reaction: Al(NO3)3(aq) + 3 KOH(aq) → Al(OH)3(s) + 3 KNO3(aq) Next, we need to calculate the number of moles of Al(NO3)3 and KOH available for this reaction using the following equation: moles (molarity)(Liters of solution) moles Al(NO3)3 (0.200 M)(0.050 L) 0.010 mol moles KOH (0.100 M)(0.200 L) 0.0200 mol Looking at the balanced equation at the top, we see that 1 mole of Al(NO3)3 needs 3 moles of KOH to react completelyTherefore 0.010 mole of Al(NO3)3 would require 0.030 mole of KOHSince only 0.0200 mol of KOH is available, then KOH would be the limiting reactant, while the Al(NO3)3 is present in excessSince the amount of Al(OH)3 produced is determined by the limiting reactant KOH, we can use dimensional analysis and equation coefficients to convert 0.0200 mole of KOH to moles of Al(OH)3 to grams of Al(OH)3We can use the following equalities to set up conversion factors: 1 mol Al(OH)3 3 mol KOH 1 mol Al(OH)3 78.00352 g Al(OH)3 [0.0200 mol KOH)/1][(1 mol Al(OH)3)/(3 mol KOH)][(78.00352 g Al(OH)3)/(1 mol Al(OH)3)] 0.5200234 g Al(OH)3 or 0.52 g Al(OH)3 rounded to two significant figures Answer: About 0.52 gram of Al(OH)3 is produced when 50.0 mL of .200 M Al(NO3)3 reacts with 200.0 mL of .100 M KOH.
