In letter A, we have to find out the number of moles of NaOH from the given volume, which must be in Liters, and also the given concentration, which is 1.00 M of NaOH
The volume found was = 5.625 mL, which in liters will be 0.00562 Liters
The formula we must use is:
n = M * V
n = 1.00 * 0.00562
n = 0.005625 moles of NaOH
In letter B, we have to calculate the moles that are in excess of HCl, the reaction that we have is:
NaOH + HCl -> NaCl + H2O
We have:
0.005625 moles of NaOH
0.0150 moles of HCl
According to the molar ratio between NaOH and HCl, we have a ratio of 1 mol of NaOH for 1 mol of HCl, therefore:
1 NaOH = 1 HCl
0.005625 NaOH = x HCl
x = 0.005625 moles of HCl
0.0150 - 0.005625 = 0.009375 moles of HCl of excess
