Hydrogen is the limiting reactant because when doing a stoichiometry equation for the reactants, hydrogen will be used completely by having a smaller yield and oxygen will be excess (7 moles to be exact)
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
2.20 M means there are 2,20 mol of NaOH in 1 000 mL of solution. We can use this proportion as a conversion factor:
Answer:
CH3COONa : CH3COOH = 1.74 : 1 (required to prepare a buffer at PH = 5)
Explanation:
Given that :
PH = 5
Calculate proportions of ethanoic acid and Sodium ethanoate to prepare Buffer solution
solution :
PH = Pka + Log ( CH3COONa / CH3COOH )
5.0 = 4.76 + Log ( CH3COONa / CH3COOH )
∴ Log ( CH3COONa / CH3COOH ) = 0.24
( CH3COONa / CH3COOH ) = 1.74
Therefore : CH3COONa : CH3COOH
1.74 : 1
