Chemical Formulae, Equations and Calculations Part 2
7 marks · HardAmmonium chloride reacts with sodium hydroxide according to the following equation:
A student adds 50.0 cm³ of 1.20 mol/dm³ sodium hydroxide solution to an excess of ammonium chloride solution in a flask at room temperature and pressure (rtp).
Calculate the number of moles of sodium hydroxide added.
Determine the maximum number of moles of ammonia gas (NH₃) that could be produced from this amount of sodium hydroxide.
Calculate the volume (in cm³) of ammonia gas produced at rtp.
The student wants to repeat the experiment but only collect 480 cm³ of ammonia gas at rtp. Calculate the minimum volume (in cm³) of 1.20 mol/dm³ sodium hydroxide solution required to produce this amount of ammonia gas.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 0.060 mol · must show Working showing cm³ to dm³ conversion and multiplication by concentration
(b) 0.060 mol · must show Recognition of 1:1 stoichiometric ratio between NaOH and NH3
(c) 1440 cm³ · must show Conversion of moles (0.060) to volume using 24,000 cm³ per mole
(d) 20.0 cm³ · must show Conversion of desired volume to moles, then solve for solution volume
Common mistakes: Using the wrong volume units conversion; Using the wrong stoichiometric ratio; Incorrectly calculating number of moles or volume
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correct calculation of number of moles of NaOH: n = (50.0 / 1000) × 1.20
M2 (b) · 1 mark · Correct use of the stoichiometric ratio (1:1) to find moles of NH3 formed = moles of NaOH used. (follow-through)
M3 (c) · 1 mark · Substitute the number of moles found in (b) into the molar volume equation (use 24,000 cm³). (follow-through)
A1 (c) · 1 mark · Correct answer: 1440 cm³ (accept in range 1435–1445); must have correct units. (follow-through)
M4 (d) · 1 mark · Convert required volume of NH₃ to moles: 480 cm³ / 24,000 cm³/mol.
M5 (d) · 1 mark · Rearrange concentration: volume = moles / concentration, use correct units. (follow-through)
A2 (d) · 1 mark · Correct calculation and final answer for volume of NaOH solution required: 20.0 cm³ (accept 19–21 cm³), with correct units. (follow-through)
Worked solution
(a) Number of moles of NaOH:
(b) Stoichiometry of the equation is 1:1: 1 mol NH4Cl reacts with 1 mol NaOH to give 1 mol NH3.
So maximum number of moles of NH₃ formed = 0.060 mol.
(c) Volume of NH₃ gas at rtp:
(d) To produce 480 cm³ of NH₃ gas:
Volume of 1.20 mol/dm³ NaOH solution required for 0.020 mol:
To 2 significant figures, minimum required volume = 20.0 cm³ (accept rounding for practicality in experiment to nearest cm³).