Question

2K(s)+2H2O(l)→2K+(aq)+2OH−(aq)+H2(g).

with the equation above,

part C

Assume that a chunk of potassium weighing 8.60 g is dropped into 400.0 g of water at 29.0 ∘C. What is the final temperature of the water if all the heat released is used to warm the water?

PART D

What is the molarity of the KOH solution prepared in part (c)?

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in....typed format...

$\Rightarrow$Answer:

1. part - ( C ):  Final temperature of water:  T_f = 54.02 ^oC
2. part - ( D ):  Molarity Of  KOH: prepared =  0.5375 M ( mol/L )

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer.

• Given:

1. ​​​​​​​balanced chemical equation: 2 K(s) + 2H₂O (l)  $\rightleftharpoons$ 2 K⁺ (aq) + 2 OH^- (aq) + H₂ (g) ------- Equation - 1
2. Mass of potassium: ( K ) = 8.60 g ( grams )
3. Mass of water: potassium is dropped into: m_w = 400.0 g ( grams )
4. Initial temperature of water: T_i = 29.0 ^oC

• ​​​​​​​Step - 1:

We know the following: molar masses:

1. molar mass of : potassium: K = 40.0 g/mol ( approx. )
2. Specific heat of liquid water:  C_w = 4.184 J / g. ^oC

• ​​​​​​​Step - 2:

​​​​​​​We know the following formula: for determining the value of standard enthalpy of reaction.

$\Rightarrow$$\Delta$H^o_rxn =   $\Sigma$ v_p x(  $\Delta$H^o_f[ products ) ] -      $\Sigma$ v_r x(  $\Delta$H^o_f [ reactants ) ] -------------------Equation - 2

$\Rightarrow$Where:

1. v_p =   total number of moles of products
2. v_r = total number of moles of reactants
3. H^o_f =   standard enthalpy of formation per mole, of products / reactants

• Step - 3:

​​​​​​​We know, the following standard enthalpy of formation: values:

1. $\Delta$H^o_f[ K(s) ] = 0.0 kJ / mol  ( kilo-joule per mole )
2. $\Delta$H^o_f[ H₂O (l) ] = - 285.8 kJ /mol
3. $\Delta$H^o_f[ K⁺ (aq) ] = - 251.2 kJ/mol
4. $\Delta$H^o_f[ OH ^- (aq) ] = - 229.9 kJ /mol
5. $\Delta$H^o_f[ H₂ (g) ] = 0.0 kJ /mol

​​​​​​​Therefore: we can plug in the values in Equation - 2, above to find the value of the standard enthalpy of reaction:

$\Delta$H^o_rxn..

• Step - 4:

​​​​​​​We know: the number of moles of potassium =  ( 8.60 g ) / ( 40.0 g/mol ) = 0.215 mol ( moles )

$\Rightarrow$ From: Equation - 1: we know: the following molar ratio:

$\Rightarrow$ molar ratio:   K : H₂O : K⁺ : OH ^- : H₂ = 1 : 1 : 1 : 1 : 0.5 ( i.e. by dividing Equation - 1: by 2.0 )

$\Rightarrow$ Therefore: if we have: 0.215 moles, of potassium ( K ) ; we will have the following moles of the compounds:

1. Number of moles of potassium (K ) = 0.215 mol  
2. Number of moles of water (H₂O ) = 0.215 mol
3. Number of moles of K⁺ = 0.215 mol    
4. Number of moles of OH ^- = 0.215 mol    
5. Number of moles of H₂ (g) = 0.1075 mol    

• ​​​​​​​Step - 5:

​​​​​​​We have the following: Equation - 2; after we plug in the values:

$\Rightarrow$$\Delta$H^o_rxn = [  (0.215 mol ) x ( $\Delta$ H^o_f [ K⁺ (aq) ] ) + (0.215 mol ) x ( $\Delta$ H^o_f [ OH ^- (aq) ]) + (0.1075 mol ) x ( $\Delta$ H^o_f [ H₂ (g) ] ) ] - [  (0.215 mol ) x ( $\Delta$ H^o_f [ K(s) ] ) + ( 0.215 mol ) x ( $\Delta$ H^o_f [ H₂O (l) ] )  

$\Rightarrow$$\Delta$H^o_rxn =   [  (0.215 mol ) x ( - 251.2 kJ/mol ) + (0.215 mol ) x ( - 229.9 kJ /mol ) + (0.1075 mol ) x (0.0 kJ/mol) ] - [  (0.215 mol ) x ( 0.0 kJ/mol ) + ( 0.215 mol ) x ( - 285.8 kJ /mol )  ]

$\Rightarrow$$\Delta$H^o_rxn = - 41.88 kJ ( i.e. the minus sign indicates , release of heat )

We know the above: heat energy: will heat up the given amount of water, in which the given mass of potassium: was dropped into.

Therefore: ..

• Step - 6:

​​​​​​​Therefore: we can:say that the amount of heat absorbed by water = $\Delta$ H^o_rxn, and using the Equation of Heat Transfer: we will get the following:

$\Rightarrow$  m_w x C_w x $\Delta$ T = ( 400.0 g ) x ( 4.184 J / g. ^oC ) x ( T_f -  29.0 ) ^oC = 41880 J

$\Rightarrow$ ( T_f -  29.0 ) ^oC =  25.02  ^oC, therefore: ...

$\Rightarrow$ Final Temperature: T_f = ( 25.02 + 29.0 ) =  54.02 ^oC

• Answer:

​​​​​​​Final temperature of water:  T_f = 54.02 ^oC

• Step - 7:

​​​​​​​Since, as we know: 0.215 moles Of  KOH: will remain dissolved in 400.0 g or 400.0 mL of Water, we can say:

$\Rightarrow$  400.0 mL of water contains:  0.215 moles Of  KOH:

$\Rightarrow$1000.0 mL of water will contain: [( 0.215 mol ) / ( 400.0 mL ) ] x ( 1000 mL ) =  0.5375 moles Of  KOH

Therefore:

$\Rightarrow$ Molarity of   0.215 moles Of  KOH: = 0.5375 M ( mol/L )