2H2 + O2 ---->2H2O
number of moles in reaction 2 mol 1 mol 2 mol
number of liters in the reaction 2*22.4 L 1*22.4 L 2*22.4L
We can see that volumes of the gases are proportional to coefficients in the reaction ( if gases are under the same conditions), so we can write
2H2 + O2 ---->2H2O
2 L 1 L 2 L
given 40 L ( 25 L) 40 L
We can see that we have excess of O2,
because if 2 L H2 are needed 1 L O2, then 40 L of H2 are needed 20 L O2.
So, limiting reactant is H2, and we will need to calculate Volume of H2O using H2.
2L H2 give 2L H2O(gas), so 40 L H2 give 40 L H2O.
Answer:
0.363 g
Explanation:
2As + 3Cl₂ → 2AsCl₃
calculate the number of moles of arsenic
Number of moles = mass/ molar mass
Number of moles = 0.150 g/ 75 g/mol
Number of moles = 0.002 mol
compare the moles of arsenic with arsenic chloride.
As : AsCl₃
2 : 2
0.002 : 0.002
Mass of arsenic chloride produced:
Mass = number of moles × molar mass
Mass = 0.002 × 181.28 g/mol
Mass = 0.363 g
The question is incomplete, complete question is;
A solution of 
is added dropwise to a solution that contains
of 
and 
and 
.
What concentration of 
is need to initiate precipitation? Neglect any volume changes during the addition.
value 
value 
What concentration of is need to initiate precipitation of the first ion.
Answer:
Cadmium carbonate will precipitate out first.
Concentration of is need to initiate precipitation of the cadmium (II) ion is 
.
Explanation:
1) 
The expression of an solubility product of iron(II) carbonate :
![K_{sp}=[Fe^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=2.10%5Ctimes%2010%5E%7B-11%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B2.10%5Ctimes%2010%5E%7B-11%7D%7D%7B1.15%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=1.826\times 10^{-9}M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D1.826%5Ctimes%2010%5E%7B-9%7DM)
2) 
The expression of an solubility product of cadmium(II) carbonate :
![K_{sp}=[Cd^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=1.80%5Ctimes%2010%5E%7B-14%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B1.80%5Ctimes%2010%5E%7B-14%7D%7D%7B0.58%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=3.103\times 10^{-12} M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D3.103%5Ctimes%2010%5E%7B-12%7D%20M)
On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.
So, cadmium carbonate will precipitate out first.
And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the concentration.
