Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl2(g)+3F2(g)→2ClF3(g) A 1.60 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 877 mmHg . Part A Identify the limiting reactant and determine the theoretical yield of ClF3 in grams.

Limiting reagent: fluorine.

Theoretical yield of ClF₃: 4.62 g.

Explanation:

Hello,

In this case, one could identify the limiting reagent by comparing the available chlorine with the consumed chlorine by the fluorine considering the ideal gas equation and the 1 to 3 respectively stoichiometric relationship between them as shown below:

In such a way, there are 0.004 excess moles of chlorine, which means that fluorine is the limiting reagent. Therefore, the theoretical yield of ClF₃ turns out:

Best regards.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"