Question

im not sure how to do the calculations for trial 1

Name La Partner Lab Section Lab 8 DATA SHEET Using Experimental Data to Verify Coefficients s Determining Mass and Verifying Reaction Product Tria 1 Mass of empty crucible + cover stor trst heating 41.73 9 Mass of empty cruce cover for second heating Final mass of empty crucible after additional heating of necessary Mass of crucible cover + MOO, -45749 Mass of crucible + cover+ H atter first heating 43.13 9 Mass of crucible + cover + K ater second heating Final mass of crucible + cover + Cl after additional heating of necessary Appearance after addition of AgNO, to KCIO, in water no reaction Appearance after addition of AgNo, to KCl in water cloud Appearance after addition of AgNo, to reaction product in water CU Calculations Show Calculations Trial 1 Show Calculations KCIO NCI S how Calculations Mass 1 Moles (mol) Moleto mole ratios Reaction coefficients whole number ratios Lab 8 • Reaction Stoichiometry 113

not sure what to do for calculations because it is needed to report experimental reaction coefficients for the decomposition of KCIO3

Answer 1

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

$\Rightarrow$Answer:

1. Mass of KClO₃ = 2.01 g; and the number of moles of KClO₃ = 0.0164 mol ( moles )
2. Mass of KCl = 2.0 g ( grams ); and the number of moles of KCl = 0.0268 mol ( moles )
3. Mass of  O₂ = 0.01 g ;  and the  number of moles of O₂ = 0.0003125 mol ( moles )
4. Molar ratio: KClO₃ : KCl : O₂ = 2 : 2 : 3

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer(s)..in...typed format...

• Given:

1. Mass of empty crucible + cover; after first heating:  m₁ = 41.73 g ( grams )
2. Mass of crucible + cover + KClO₃ : m₂ = 43.74 g ( grams )
3. Mass of crucible + cover + KCl , after first heating: m₃ = 43.73 g ( grams )

• Step - 1:

​​​​​​​We know the following:

1. Molar mass of  KClO₃ = 122.55 g/mol
2. Molar mass of KCl = 74.55 g/mol  
3. Molar mass of O₂ = 32.0 g/mol

• ​​​​​​​Step - 2:

​​​​​​​From the given data: as we can see: we can calculate the following:

$\Rightarrow$ Mass of  KClO₃ = m₂ - m₁ =  ( 43.74 g - 41.73 ) = 2.01 g ( grams )

$\Rightarrow$ Therefore: number of moles of KClO₃ = ( 2.01 g ) / ( 122.55 g/mol ) = 0.0164 mol ( moles )

• Step - 3:

​​​​​​​Using the given data: we can calculate:

$\Rightarrow$ Mass of KCl =   m₃ -   m₁ =  ( 43.73 - 41.73 ) = 2.0 g ( grams )

$\Rightarrow$ Therefore: number of moles of KCl = ( 2.0 g ) / ( 74.55 g/mol ) = 0.0268 mol ( moles )

• Step - 4:

​​​​​​​Using the given data: we can also calculate the following:

$\Rightarrow$ Mass of oxygen: i.e. O₂ = m₂ - m₃ =   ( 43.74 - 43.73 ) = 0.01 g

$\Rightarrow$ Therefore: number of moles of O₂ = ( 0.01 ) / ( 32.0 g/mol ) = 0.0003125 mol ( moles )

• Step - 5:

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

$\Rightarrow$ 2 KClO₃ (s) $\rightleftharpoons$ 2 KCl (s) + 3 O₂ (g) -------------------------------------Equation - 1

$\Rightarrow$ Therefore: the molar ratio: will be the following:

  KClO₃ : KCl : O₂ = 2 : 2 : 3

[ Note: The above answer: has been derived, not on the basis of the experimental data: but, rather it is based on the basis of the stoichiometry, and balanced chemical equation: of the reaction. ]