Question #7e002

1 Answer
Apr 28, 2016

Answer:

#color(red)(2)NaCl + H_2SO_4 = Na_2SO_4+color(red)(2)HCl#

Explanation:

Since we have #Na_color(red)(2)SO_4# on the right side
the left side must have a multiple of #color(red)(2)# #NaCl#'s

Trying the minimum of #color(red)(2)NaCl# on the left implies we will need to have a multiple of #color(red)(2)color(white)(" ")Cl#'s in #HCl# on the right;

but the #H_color(red)(2)# in the #H_2SO_4# on the left provides the #color(red)(2) H#'s needed for #color(red)(2)HCl# to work out.