Question

queous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. Suppose 6.93 g of hydrochloric acid is mixed with 2.4 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits. g

Answer 1

Answer:

Explanation:

NaOH   +   HCl   =   NaCl   +   H₂O

1 mole      1 mole                     1 mole

6.93 g of HCl = 6.93 / 36.5 mole = .19  mole

2.4 g of NaOH = 2.4 /40 mole of NaOH = .06 mole

.06 mole of NaOH  is less than .19 mole of HCl so

limiting reagent is NaOH

.06 mole of NaOH will react with .06 mole of HCl to give .06 mole of water .

water formed by given mass of reactants

= .06 moles of water

= .06 x 18 gram of water

= 1.08 g of water .

= 1.1 g of water .