Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O . Suppose 23.7 g of hydrochloric acid is mixed with 40. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.
HCl + NaOH = NaCl + H2O
first we have to calculate the moles of HCl and NaOH so from this we can find the limiting reagent in this reaction
now moles of HCl = weight of HCl/ molecular weight of HCl
= 23.7/36.5 = 0.6493
similiarly moles of NaOH = weight of NaOH/molecular weight of NaOH
= 40/40 = 1 moles
so now in this reaction limiting rewagent is HCl (0.6493 moles) reacted with 1 mole of NaOH
thearticall yield of any reaction is = molecular weight of anticipated product x weight of limiting reagent/molecular weight of limiting reagent
now weight of can be produced in this reaction is = 18 x23.7/36.5
= 11.68 gm
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