If all could be answered. Each question can be subtracted from how many questions I have left to ask. NAMES:SECTION:Trial #1Trial #21. Mass of the filter paper, g08330862Mass of filter paper and precipitate, g4.5314.5642. Chemical equation:Al(NO:)s(a)NAOH(a)->3. N° moles Al(NO3)3(aq), mol4. N° moles NAOH(aq), mol5. Limiting reactant6. Reactant in excess7. Theoretical yield8. Percent yieldThinking critically:1) Explain how it is possible to start with equal volumes and equal concentrations of bothreactants, and end up with one of them being used up, and the other being in excess.2) Is the percent yield supposed to be 100%? What could explain the discrepancy?

Hello. There are multiple questions posted together. The first question has been solved. To receive solutions for the remaining parts, please repost the whole question and mention the parts to be solved.Since the data on the obtained yield or initial amounts of reactants are not provided, the limiting reagent and reactant in excess cannot be predicted. 1. The chemical reaction taking place is, Al(NO3)3 (aq) + 3 NaOH (aq)→Al(OH)3 (s) +3 NaNO3 (aq) From the balanced equation of the reaction it can be inferred that, 1 mole of aluminium nitrate reacts with 3 moles of sodium hydroxide to give one mole of aluminium hydroxide (precipitate) and 3 moles of sodium nitrate (soluble in water). Since the reactants react with one another in a 1:3 ratio (Al(NO3)3:NaOH), if equal amounts of reactants (1:1) are taken, then some moles of unreacted aluminium nitrate will be left out, as only 1x moles of aluminium nitrate will have reacted with 3x moles of NaOH. The amount of reagent used up or left over is dependent on the mole ratio of reactants reacting with one another. Thus by taking equal amounts (Volume and concentration) of both reactants, some amounts of one of the reactant will be in excess when the reactants do not react in a 1:1 mole ratio.