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
May 15, 2017

Answer:

#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"#