# Question 7ee1d

May 17, 2016

${n}_{S C {l}_{2}} = 1.25 \setminus m o l$

#### Explanation:

$\text{S + Cl"_2 -> " SCl"_2 " equation 1}$
${m}_{1} \text{ " m_1^'}$

$\text{2S + Cl"_2->" S"_2"Cl"_2 " equation 2}$
${m}_{2} \text{ } {m}_{2}^{'}$

$\text{Mass of Cl"_2 " used in equation 1}$

${m}_{1}^{'} = {m}_{1} \times \left(1 \setminus m o l . {\text{S")/(32.0066\ g" S")xx (1\ mol " Cl"_2)/(1\ mol" S")xx(70.91\ g" Cl"_2)/(1\ mol" Cl}}_{2}\right)$

${m}_{1}^{'} = 2.22 . {m}_{1}$

$\text{Mass of Cl"_2 " used in equation 2}$

${m}_{2}^{'} = {m}_{2} \times \left(1 \setminus m o l . {\text{S")/(32.0066\ g" S")xx (1\ mol " Cl"_2)/(2\ mol" S")xx(70.91\ g" Cl"_2)/(1\ mol" Cl}}_{2}\right)$

${m}_{2}^{'} = 1.11 . {m}_{2}$

The total mass of sulfur used is 50.0 g and the total mass of chlorine used 100. g

${m}_{1} + {m}_{2} = 50.0 \text{ ( Eq. 2)}$

${m}_{1}^{'} + {m}_{2}^{'} = 100. \text{ (Eq. 3)}$.

Now, you have a system of two equations with two unknowns.

$\text{(Eq.3)" rArr 2.22*m_1 + 1.11*m_2=100. " (Eq.4)}$

$\text{(Eq. 2)" rArr m_2= 50.0 - m_1 " (Eq.5)}$

$\text{(Eq.4) + (Eq.5) } \Rightarrow$

$2.22 \cdot {m}_{1} + 1.11 \cdot \left(50.0 - {m}_{1}\right) = 100.$

Solve for ${m}_{1}$

${m}_{1} = 40.1 \setminus g$

${m}_{2} = 9.9 \setminus g$

 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"))#

${n}_{S C {l}_{2}} = 1.25 \setminus m o l$