Where is the link can’t do any work without it
        
             
        
        
        
Answer:
Where is the chemical equations?
 
        
                    
             
        
        
        
it is already balanced
REACTANTS 
Barium sulfide (BaS) + platinum (Ii) fluoride 
PRODUCT
Barium fluoride (BaF2) + Cooperite (PtS)
Hope this answer helps you dear! take care
 
        
             
        
        
        
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is 
 of butane reacted
    of butane reacted 
Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol 
                      = 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol 
                                         = 99.3 g of CO₂ 
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
 
        
             
        
        
        
Answer:
Primero debes usar los gramos de co2 y luego buscar su peso molecular, luego de eso usar la relación de moles entre CO2 y H2O y por último buscar el pm del H2O pata ver cuantos gramos de produce.
Explanation: