2 H2(g) + S2(g) equilibrium reaction arrow 2 H2S(g) At a certain temperature, Kc = 1.30 ✕ 1010 for the reaction above. Calculate the equilibrium concentration of H2S if [H2] = 0.00400 M and [S2] = 0.00270 M at equilibrium in a container.

The equilibrium concentration of hydrogen sulfide is 23.7 M

Explanation:

For the given chemical equation:

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the negative value of 'x' because concentration cannot be negative

Hence, the equilibrium concentration of hydrogen sulfide is 23.7 M

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the answer is c. nacl