The following data was collected when a reaction was performed experimentally in the laboratory. reaction data reactants products al(no3)3 nacl nano3 alcl3 4 moles 9 moles ? ? determine the maximum amount of alcl3 that was produced during the experiment. explain how you determined this amount.

First thing you need to do is write the chemical equation and check to see if it is balancedAl(NO₃)₃ + NaCl ⇒ NaNO₃ + AlCl₃The left hand side requires 3 more chloride ions, while the right hand side needs 3 nitrate ions to be balanced. Add a coefficient of 3 on NaCl and NaNO₃ to get a balanced equation.Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃The ratio of Al(NO₃)₃: NaCl is 1:3, which means 1 mol of Al(NO₃)₃ reacts with 3 moles of NaCl ˣ =Clearly, NaCl is the limiting reactant since according to the ratio 1:3, 4 mol Al(NO₃)₃ should react with 12 mol NaCl, but the reaction data has 9 mol of NaCl available. Use the limiting reactant moles to find the product NaNO₃ formed.The molecular weight of NaNO₃ is 84.9965 g/mol and thus mass of NaNO₃ formed is mass = (9 moles NaNO₃)(84.9965 g/mol NaNO₃)= 764.97g

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

20 drops hope this