What is the limiting reactant when 7.81 of HCl reacts with 5.24 g of NaOH to produce NaCl and H_2O?

1 Answer
anor277
May 15, 2017

NaOH

Explanation:

For the reaction,

NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)........

Clearly, there is 1:1 stoichiometry.

"Moles of HCl"=(7.81*g)/(36.5*g*mol^-1)=0.214*mol.

"Moles of NaOH"=(5.24*g)/(40.0*g*mol^-1)=0.131*mol.

Clearly, HCl is in excess. Note that HCl is a room temperature gas (that has some great solubility in water). Normally, an aqueous solution of a known concentration would be proposed for the acid, and you would access the molar quantity of HCl by the relation:

"Concentration"="Moles of HCl"/"Vollume of solution"

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