Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl2(g)+3F2(g)→2ClF3(g) A 2.05 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 730 mmHg .

2.4 grams of ClF3

Explanation:

First let us determine the moles of Cl2 and F2,

Cl2 = (  ( 337 )( 2.05 L ) / ( 0.082 )( 298 K ) ) * ( 1 atm / 780 ),

Cl2 = ( 690 / 24.436 ) * ( 1 / 780 ),

Cl2 = ( About ) 0.036 moles of Cl2

F2 = ( ( 729 )( 2 L ) / ( 0.082 )( 298 K ) ) * ( 1 atm / 780 ),

F2 = ( 1458 / 24.436 ) * ( 1 / 780 )

F2 = ( About ) 0.078 moles of F2

Now let us identify the limiting reactant, considering the ratio between ClF3 and Cl2 / F2. In this case F2 is the limiting reactant, as it forms a smaller molar ratio;

The theoretic yield is thus performed with the limiting reactant F2,

0.078 * ( 2 / 3 ) * ( 92.45 / 2 ) = ( About ) 2.4 grams of ClF3

the answer is barium and copper bromide

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o