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

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