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2 years ago

How many grams of H,O are needed to produce 13 g of NaOH? 2 Na,O2 + 2 H20-4 NaOH + O2

Chemistry

1 answer:

fgiga [73]2 years ago

3 0

Answer:

This is a typical stoichiometry question.To answer this question you want to get a relationship between

O and NaOH.

So you can get a relationship between the moles of

and moles of NaOH by the concept of stoichiometry.

O +

O ----------------> 2 NaOH.

According to above balanced equation we can have the stoichiometry relationship between

O and NaOH. as 1:2

It means 1 moles of

O is required to react with one mol of

O to produce 2 moles of NaOH.

in terms of mass 1 mole of

O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.

62 g of

O produces 80 g of NaOH.

1g of NaOH is produced from 62/80 g of

1.6 x

g of NaOH will require 62 x 1.6 x

g / 80 of

124g of

Explanation:

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From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

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Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

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