Answer: 2.8 moles
Explanation:
The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.
Na2O + H2O --> 2NaOH
1 mole of H2O = 2 moles of NaOH
Let Z moles of H2O = 5.6 mole of NaOH
To get the value of Z, cross multiply
5.6 moles x 1 mole= Z x 2 moles
5.6 = 2Z
Divide both sides by 2
5.6/2 = 2Z/2
2.8 = Z
Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH
Answer:
2.58 L
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Explanation:
Write the balance chemical equation ,
initial concenration of 
lets assume that degree of dissociation=
concenration of each component at equilibrium:
![[SO_2Cl_2] = 0.1-0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha)
![[SO_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2%5D%20%3D%200.1%5Calpha)
![[Cl_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BCl_2%5D%20%3D%200.1%5Calpha)
as is very small then we can neglect 
therefore ,
Eqilibrium concenration of ![[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha%20%3D%200.1-0.1%5Ctimes%200.00173)
91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Explanation:
2NaN3======2Na+3N2
This is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.
From the equation:
2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.
In the question moles of nitrogen produced is given as 2.104 moles
so,
From the stoichiometry,
3N2/2NaN3=2.104/x
= 3/2=2.104/x
3x= 2*2.104
= 1.4 moles
So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.
From the formula
no of moles=mass/atomic mass
mass=no of moles*atomic mass
1.4*65
= 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
