Question

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`?

Answer 1

`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`