54g ag *(108mol ag/1 g ag) =5832mol ag
C6H12O6 +6O2 —> 6CO2 + 6H2O
Answer:
Explanation:
mole of NaOH present = molarity x volume
= 1.0 X 0.05 = 0.05 mole
<em>Recommended mole of HCl </em>= 1.1 x 0.05 = 0.055
<em>Mole of HCl carelessly added by Jacob </em>= 1.1 x 0.04 = 0.044
From the equation of reaction:
HCl + NaOH ----> NaCl + H2O
The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>
The mole of HCl added by Jacob (0.044) is short of the recommended amount (0.055) and also short of the amount required for a neutral endpoint (0.05). <u>This means that the endpoint will have an excess amount of NaOH and as such, basic instead of the desired acidic endpoint.</u>
