# What mass of SbCl_3 can be produced from the reaction of 3.570 L of chlorine at STP reacting with an excess of antimony in the reaction Sb+Cl_2 -> SbCl_3?

Jun 9, 2016

$24.1 \setminus g$

#### Explanation:

First, start by writing a balanced chemical equation for the reaction

$2 S b + 3 C {l}_{2} \to 2 S b C {l}_{3}$

Then find the number of moles of chlorine gas involved in the chemical reaction use the ideal gas equation.

$P \cdot V = n \cdot R \cdot T$

$P = 1.00 \setminus a t m$

$T = 273 \setminus K$

$V = 3.570 \setminus L$

$R = 0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot \setminus {K}^{-} 1$

Solve for the number of moles $n$.

$n = \frac{P \cdot V}{R \cdot T}$

$n = \frac{1.00 \setminus a t m \times 3.570 \setminus L}{0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot \setminus {K}^{-} 1 \times \setminus 273 \setminus K}$

$n = \frac{1.00 \setminus \cancel{a t m} \times 3.570 \setminus \cancel{L}}{0.0821 \setminus \cancel{L} \cdot \cancel{a t m} \cdot m o {l}^{-} 1 \cdot \cancel{{K}^{-} 1} \times \setminus 273 \setminus \cancel{K}}$

$n = 1.59 \times {10}^{-} 1 \setminus m o l . C {l}_{2}$

The molar mass of $S b C {l}_{3} = 228.11 \setminus g . m o l {.}^{-} 1$

Using the the number of moles of the $C {l}_{2}$ the mass of a $S b C {l}_{3}$ 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

$1.59 \times {10}^{-} 1 \setminus \cancel{m o l . C {l}_{2}} \times \frac{2 \setminus \cancel{m o l . S b C {l}_{3}}}{3 \cancel{m o l . C {l}_{2}}} \times \frac{228.11 \setminus g \setminus S b C {l}_{3}}{1 \setminus \cancel{m o l . S b C {l}_{3}}}$

$= 24.1 \setminus g$