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#?

1 Answer
Jun 9, 2016

Answer:

#24.1 \g#

Explanation:

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#