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