Sodium hydroxide reacts with carbon dioxide as follows: 2NaOH(s)+CO2(g) yields NaCO3(s)+H2O(l)?

a) NaOH is the limiting reagent ( 2 moles are needed=
b) 1.85/2 =0.925 moles Na2CO3
c) moles CO2 needed = 1.85/2 = 0.925
moles CO2 in excess = 1.00 - 0.925=0.075

start up with a balanced equation first 2 NaOH + CO2 ---------> Na2CO3 + H2O a) 2 moles of NaOH react with a million mole of CO2, you have in basic terms a million.80 5 moles of NaOH, so it is the restricting reagent b) i'm assuming you're using the comparable a million.80 5 and a million mole ratios. Now 2 moles of NaOH produce a million mole of Na2CO3, then a million.80 5 moles will produce a million.80 5/2 = 0.925 moles of sodium carbonate. while you're touching on in basic terms the equation, then 2 moles will produce one mole of sodium carbonate c) 2 moles of NaOH react with a million mole of CO2, then a million.80 5 moles will react with a million.80 5/2 = 0.925 moles of CO2, so extra left would be a million - 0.925 = 0.0.5 moles of CO2 would be left in the back of. you additionally can arrive on the respond from the records in b above. One mole of CO2 produces a million mole of sodium carbonate, so 0.925 moles of carbonate will use 0.925 moles of CO2 and then a million - 0.925 = 0.0.5 moles of CO2 left in the back of There you have it