Question

Question 1 include A, B & C

I want to produce nitric acid, HNO3 using the unbalanced reaction below. In the lab I have 250ml of a 0.50M solution of barium nitrate. Ba(NO3)2 + H2S →BaS + HNO3. A. Balance the reaction?

Answer 1

Following is the - complete Answer -&- Explanation: for the first question (i.e. Question - 1 ), of the given: Question Set.....in....typed format...

$\Rightarrow$Answer:

1. part - (A):   Ba(NO₃)₂ + H₂S $\rightleftharpoons$ BaS + 2 HNO₃ ( balanced equation of reaction )
2. part - (B): Number of moles of barium nitrate; Ba(NO₃)₂ =  0.125 mol ( moles )
3. part -(C): Number of moles of nitric acid ( HNO₃ ) possible to be produced = 0.25 mol ( moles )
4. part - (D): Final concentration of nitric acid; HNO₃ =   1.0 M ( mol/L )

$\Rightarrow$Explanation:

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

• Given:

1. ​​​​​​​We have the following compounds: barium nitrate ( Ba(NO₃ )₂ ); hydrogen sulfide ( H₂S ); barium sulfide

( BaS ) ; and:  nitric acid ( HNO₃ )

2. Volume of the solution: of barium nitrate : V_ba(no3)2 = 250.0 mL = 0.250 L ( Liters )
3. Molar concentration of the solution:of barium nitrate: M_ba(no3)2 = 0.50 M ( mol/L )
4. Unbalanced chemical equation: Ba(NO₃)₂ + H₂S $\rightarrow$ BaS + HNO₃--------------------Equation - 1

• ​​​​​​​Step - 1:

​​​​​​​If we choose to balance Equation - 1: we would get the following:

$\Rightarrow$balanced chemical equation:

   Ba(NO₃)₂ + H₂S $\rightleftharpoons$ BaS + 2 HNO₃-----------------------------Equation - 2

• Step - 2:

​​​​​​​$\Rightarrow$ Number of moles of barium nitrate:=  V_ba(no3)2 x  M_ba(no3)2 = ( 0.250 L ) x ( 0.50 mol/L ) = 0.125 mol ( moles )

• Step - 3:

​​​​​​​We have the following: balanced chemical equation:

$\Rightarrow$   Ba(NO₃)₂ + H₂S $\rightleftharpoons$ BaS + 2 HNO₃-----------------------------Equation - 2

Therefore:

$\Rightarrow$molar ratio:    Ba(NO₃)₂ :  HNO₃ = 1 : 2   

$\Rightarrow$ We know: number of moles of barium nitrate, Ba(NO₃)₂ =  0.125 mol ( moles )

Therefore:

$\Rightarrow$ Using the above: molar ratio: we would get:

$\Rightarrow$ Number of moles of nitric acid ( HNO₃ ) produced = 2 x ( 0.125 mol ) =  0.25 mol ( moles )

• Step - 4:

​​​​​​​$\Rightarrow$ We know, the volume of barium nitrate : used = 250 mL = 0.250 L ( Liters )

$\Rightarrow$ Therefore: volume of the: final solution: = 0.250 L ( Liters )

So, we can say the following: i.e.

$\Rightarrow$ 0.250 L  ( Liters ) of the final solution: would contain:  0.25 mol ( moles) of  HNO₃

$\Rightarrow$1.0 L  ( Liters ) of the final solution: would contain: (0.25 mol ) / ( 0.250 L ) = 1.0 mol ( moles) of  HNO₃

$\Rightarrow$Molar concentration, of final solution: = 1.0 mol / L =  1.0 M   ( Answer )

$\Rightarrow$ Final concentration of nitric acid : HNO₃ = 1.0 M ( mol/L )