## General Chemistry: Principles and Modern Applications (10th Edition)

It is necessary to have 0.0539L of HCl to neutralize this quantity of $NH_3$
1. Find the volume necessary for the neutralization: $1 * {C_1 * V_1} = 1 * {C_2 * V_2}$ $0.265* 1.25= 6.15 * V_2$ $0.33125 = 6.15 * V_2$ $V_2 = 0.05386L$ of HCl.