Question #7ee1d

1 Answer
May 17, 2016

n_(SCl_2)=1.25 \ mol

Explanation:

"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