An excess of sodium hydroxide is treated with 1.1 L of dry hydrogen chloride gas measured at STP. What is the?

Question

An excess of sodium hydroxide is treated with 1.1 L of dry hydrogen chloride gas measured at STP. What is the mass of sodium chloride formed?

Answer 1

STP is 1 atm, 273 K.

moles of HCl = PV/RT = 1 atm * 1.1 L/ (0.08206 * 273K) = 0.049 moles

Since HCl and NaOH is a 1:1 reaction, 1 mole HCl gives 1 mole NaCl, so you get 0.049 moles NaCl (MW 58.5), or 2.87g.

Answer 2

PV=nRT
(1atm)(1.1L)=n(.0821Latmmol^-1K^-1)(273K)
n=0.0491mol HC
HCl+NaOH---->NaCl+HOH
n(NaCl)=0.0491mol
m(NaCl)=0.0491*58.5=2.87 gr.