#"S + Cl"_2 -> " SCl"_2 " equation 1"#
#m_1" " m_1^'" #
#"2S + Cl"_2->" S"_2"Cl"_2 " equation 2"#
#m_2" " m_2^'#
#"Mass of Cl"_2 " used in equation 1"#
# m_1^'= m_1xx (1\ mol."S")/(32.0066\ g" S")xx (1\ mol " Cl"_2)/(1\ mol" S")xx(70.91\ g" Cl"_2)/(1\ mol" Cl"_2)#
#m_1^'= 2.22.m_1#
#"Mass of Cl"_2 " used in equation 2"#
# m_2^'= m_2xx (1\ mol."S")/(32.0066\ g" S")xx (1\ mol " Cl"_2)/(2\ mol" S")xx(70.91\ g" Cl"_2)/(1\ mol" Cl"_2)#
#m_2^'= 1.11.m_2#
The total mass of sulfur used is 50.0 g and the total mass of chlorine used 100. g
#m_1 + m_2= 50.0 " ( Eq. 2)"#
#m_1^' + m_2^'= 100. " (Eq. 3)"#.
Now, you have a system of two equations with two unknowns.
#"(Eq.3)" rArr 2.22*m_1 + 1.11*m_2=100. " (Eq.4)"#
#"(Eq. 2)" rArr m_2= 50.0 - m_1 " (Eq.5)"#
#"(Eq.4) + (Eq.5) " rArr #
#2.22*m_1 + 1.11*(50.0- m_1)=100. #
Solve for #m_1#
#m_1= 40.1\ g#
#m_2=9.9\ g#
# n_(SCl_2)=40.1\ g " S"xx (1\ mol."S")/(32.0066\ g" S")xx (1\ mol " SCl"_2)/(1\ mol" S")#
# n_(SCl_2)=40.1\ cancel(g" S")xx (1\ cancel (mol."S"))/(32.0066\ cancel(g" S"))xx (1\ mol " SCl"_2)/(1\ cancel(mol" S"))#
# n_(SCl_2)=1.25 \ mol#
