Answer to Question #303600 in Chemistry for David

Question

30 cm³ of 0.1 M Al (NO₃)₃ SOLUTION IS RECTED WITH 100cm³ of 0.15M of NaOH solution. Which reactant is in excess, and by now much ?

• A. NaOH solution by 70 cm³
• B. NaOH solution, by 60 cm³
• C. NaOH solution by 40 cm³
• D. Al(NO₃)₃ solution by 20 cm³
• E. Al(NO₃)₃ solution by 10 cm³

Answer 1

Al(NO₃)₃ + 3NaOH = 3NaNO₃ + Al(OH)₃

C_M = n / V

n = C_M x V

n (Al(NO₃)₃) = 0.1 x 0.03 = 0.003 mol

n (NaOH) = 0.15 x 0.1 = 0.015 mol

According to the eqation, the required n (NaOH) = 3 x n (Al(NO₃)₃) = 3 x 0.003 = 0.009 mol

Therefore, n (NaOH) is in excess of 0.015 - 0.009 = 0.006 mol

V (NaOH)_excess = 0.006 / 0.15 = 0.04 L = 40 ml

Answer: C. NaOH solution by 40 cm³