The question is incomplete. The complete question is :
Hydrogen is manufactured on an industrial scale by this sequence of reactions:
The net reaction is :
Write an equation that gives the overall equilibrium constant 
in terms of the equilibrium constants 
and 
. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.
Solution :
![$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%3D%20%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%24)
...............(1)
![$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_2%20%3D%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
...................(2)
![$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
On multiplication of equation (1) and (2), we get
![$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%5Ctimes%20K_2%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5Ctimes%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
![$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K_1K_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
.................(4)
Comparing equation (3) and equation (4), we get
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
Answer: The mass of 
and 
produced are 336.6 g and 183.6 g respectively.
Explanation:
The combustion reaction between propane and oxygen leads to formation of carbon dioxide and water.
Law of Conservation of mass states that the mass will remain constant for a balanced equation. This is carried out when the total number of atoms on reactant side is same as the total number of atoms on the product side. Thus the equation must be balanced.
a) 1 mol of propane produces = 3 moles of
Thus 2.55 mol of propane produces = ![\frac{3}{1}\times 2.55=7.65 moles of [tex]CO_2](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B1%7D%5Ctimes%202.55%3D7.65%20moles%20of%20%5Btex%5DCO_2)
mass of 
b) 1 mol of propane produces = 4 moles of
Thus 2.55 mol of propane produces = ![\frac{4}{1}\times 2.55=10.2 moles of [tex]H_2O](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B1%7D%5Ctimes%202.55%3D10.2%20moles%20of%20%5Btex%5DH_2O)
mass of 
The mass of and produced are 336.6 g and 183.6 g respectively.
<span>every column on the table represents a family that react similarly with other <span>elements.</span></span>
