Moles of water formed in the reaction: Given 50mL of HCl 2.0M and 55mL of NaOH 2

Question

Moles of water formed in the reaction:

Given 50mL of HCl 2.0M and 55mL of NaOH 2.013M, calculate the moles of water formed in the reaction of HCl+NaOH. I calculated .0998 moles, is this correct? I found the moles of HCl to be .1 and the moles of NaOH to be .1107, and from these I found the grams to be 3.64 grams of HCl (.1 moles x 36.46 g/mol) and 4.43 grams of NaOH (.1107 moles x 39.997 g/mol). I then did a limiting reagent problem and got .0998 moles from HCl, and .11075 moles from NaOH, so .0998 moles is correct? Can someone check my math here? Thank you!

Explanation / Answer

The balanced chemical equation is

HCl (aq) + NaOH (aq) --------> NaCl (aq) + H2O (l)

As per the stoichiometric equation,

1 mole HCl = 1 mole NaOH = 1 mole H2O.

Find out the moles of HCl and NaOH as below.

Moles HCl = (volume of HCl in L)*(concentration of HCl in mol/L) = (50.0 mL)*(1 L/1000 mL)*(2.0 M)*(1 mol/L/1 M) = 0.1 mole.

Moles NaOH = (volume of NaOH in L)*(concentration of NaOH in mol/L) = (55.0 mL)*(1 L/1000 mL)*(2.013 M)*(1 mol/L/1 M) = 0.110715 mole.

Since HCl and NaOH react on a 1:1 molar ratio and we find we have fewer moles of HCl as compared to NaOH, hence, HCl is the limiting reactant (ans).

Water, H2O is a product of the reaction and the yield of the product is governed by the mole(s) of the limiting reactant taken. Since HCl and H2O have a 1:1 molar ratio, hence,

Moles of H2O produced = moles of HCl taken = 0.1 mole.

Molar mass of H2O = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol; therefore, mass of H2O produced = (0.1 mole)*(18.0154 g/mol) = 1.80154 g 1.80 g (ans).

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