Question

In: Chemistry

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

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!

Solutions

Expert Solution

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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