CO(g) +2H2--->CH3OH
2.50g H2*1mol/2g=1.25 mol H2
30.0L CO*1mol/22.4L=1.34 mol CO,
according to reaction 1 mol CO needs 2 mol H2,so 1.34 mol CO need 2.68 mol H2, so 1) limiting teactant is H2 (H)
2)1.25 mol CH3OH will be produced, 1.25 mol*32g/mol=40.0 g CH3OH
3) 1.25 mol H2 needs 0.625 g CO
1.34-0.625=0.715 g CO leftover
Answer:
The value is 
Explanation:
From the question we are told that
The concentration of
is 
The solubility product constant for
is 
The stability constant for
is 
Generally the dissociation reaction for NiS is
Generally the formation reaction for
is

Combining both reaction we have

Gnerally the equilibrium constant for this reaction is

=>
=> 
Generally the I C E table for the above reaction is

initial [ I] 0.091 0 0
Change [C] -4x +x + x
Equilibrium [E ] 0.091 - 4x x x
Here is x is the amount in term of concentration that is lost by
and gained by
and 
Gnerally the equilibrium constant for this reaction is mathematically represented as
![K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}](https://tex.z-dn.net/?f=K_c%20%20%3D%20%20%5Cfrac%7B%5BNi%20%28CN%29_4%5E%7B2-%7D%20%5D%20%5BS%5E%7B2-%7D%20%5D%20%7D%7B%20%5BCN%5E%7B-%7D%5D%5E4%7D)
=> ![3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}](https://tex.z-dn.net/?f=3.0%2A10%5E%7B12%7D%20%3D%20%20%5Cfrac%7Bx%20%2A%20%20x%7D%7B%20%5B0.091%20-%204x%20%5D%5E4%7D)
=> ![3.0*10^{12}* [0.091 - 4x ]^4 = x^2](https://tex.z-dn.net/?f=3.0%2A10%5E%7B12%7D%2A%20%20%5B0.091%20-%204x%20%5D%5E4%20%3D%20x%5E2)
=> ![[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}](https://tex.z-dn.net/?f=%5B0.091%20-%204x%20%5D%5E4%20%3D%20%20%5Cfrac%7Bx%5E2%7D%7B3.0%2A10%5E%7B12%7D%7D)
=> ![[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}](https://tex.z-dn.net/?f=%5B0.091%20-%204x%20%5D%20%3D%20%5Csqrt%5B4%5D%7B%20%5Cfrac%7Bx%5E2%7D%7B3.0%2A10%5E%7B12%7D%7D%7D)
=> ![[0.091 - 4x ] = \frac{\sqrt{x} }{1316}](https://tex.z-dn.net/?f=%5B0.091%20-%204x%20%5D%20%3D%20%5Cfrac%7B%5Csqrt%7Bx%7D%20%7D%7B1316%7D)
=> 
Square both sides

=> 
=> 
Solving using quadratic equation
The value of x is 
Hence the amount in terms of molarity (concentration) of
and
produced at equilibrium is
it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is 
So the molar solubility of nickel(II) sulfide at equilibrium is

Answer:
48.075g(or 48g in correct sig figs)***
Explanation:
=48.075g
*64.1g is the mass of SO2 which is calculated by simply taking the mass of sulfur and oxygen(but doubling it since there are two oxygens) and adding them together(32.1+2x16.0)
**btw the mol units cancel because of dimensional analysis in case anyone was wondering why
***if your teacher is like mine and specifically wants your answer in correct sig figs, use the answer in parentheses as the original problem only has 2 sig figs