Posted by **ANNA-MARIE** on Tuesday, April 19, 2011 at 6:08pm.

When SbCl3(g) (4.867 mol) and 0.01217 mol/L of Cl2(g) in a 400.0 L reaction vessel at 958.0 K are allowed to come to equilibrium the mixture contains 0.008586 mol/L of SbCl5(g). What concentration (mol/L) of SbCl3(g) reacted?

SbCl3(g)+Cl2(g) = SbCl5(g)

Unit Conversions:

K = C + 273

Molar Mass (g/mol)

SbCl3(g) 228.16

Cl2(g) 70.906

SbCl5(g) 299.06

## Answer this Question

## Related Questions

- Chemistry - When F2(g) (0.02028 mol/L) and 0.02028 mol/L of Cl2(g) in a 460.0 L ...
- CHEMISTRY - When NO(g) (0.06623 mol/L) and 422.6 grams of Cl2(g) in a 180.0 L ...
- chemistry(just need to clarify, no need to solve) - When a sample of NO2(g) (5....
- Chemistry - Equilibrium Constant - Determine the value of the equilibrium ...
- Chemistry - When a sample of I(g) (18.15 mol) is placed in 59.00 L reaction ...
- Chemistry - When a sample of I2(g) (0.07249 mol/L) is placed in 130.0 L reaction...
- college chemistry - When a sample of HI(g) (5.825 mol) is placed in 320.0 L ...
- Chemistry - When CO2(g) (0.118 mol/L) and H2(g) (3.90 mol) in a 33.0 L reaction ...
- Chemistry - When CO2(g) (0.009242 mol/L) and 0.009242 mol/L of H2(g) in a 460.0 ...
- chemistry - When CO(g) (38.11 grams) and 96.48 grams of Cl2(g) in a 140.0 L ...

More Related Questions