First, start by writing a balanced chemical equation for the reaction
#2Sb + 3Cl_2 -> 2SbCl_3#
Then find the number of moles of chlorine gas involved in the chemical reaction use the ideal gas equation.
#P*V=n*R*T#
#P = 1.00 \ atm#
#T = 273 \ K#
#V = 3.570\ L#
# R = 0.0821 \ L * atm* mol^-1*\ K^-1#
Solve for the number of moles #n#.
#n=(P*V)/(R*T)#
#n=(1.00 \ atm xx 3.570 \ L)/(0.0821 \ L * atm* mol^-1*\ K^-1xx \273 \ K) #
# n=(1.00 \ cancel(atm)xx 3.570 \ cancel( L))/(0.0821 \ cancel(L)* cancel( atm)* mol^-1*cancel(K^-1)xx \273 \ cancel(K))#
#n= 1.59xx10^-1\ mol. Cl_2#
The molar mass of #SbCl_3 = 228.11 \ g.mol.^-1#
Using the the number of moles of the #Cl_2# the mass of a #SbCl_3# could be determined.
#1.59xx10^-1 \ mol. Cl_2 xx (2\ mol. SbCl_3)/(3 \ mol. Cl_2)xx (228.11 \ g \ SbCl_3) /( 1 \ mol. SbCl_3#
#1.59xx10^-1 \ cancel(mol. Cl_2) xx (2\ cancel (mol. SbCl_3))/(3 cancel( mol. Cl_2)) xx (228.11 \ g \ SbCl_3) /( 1 \ cancel(mol. SbCl_3))#
#= 24.1\ g#