IB Chemistry HL 2.1 Amount of Chemical Change Questions
Practise IB Chemistry HL 2.1 with HL questions on balanced equations, mole calculations, limiting reactants, yield and atom economy.
- Syllabus
- First assessment 2025
- Course
- Chemistry HL
- Level
- HL
Practise IB Chemistry HL 2.1 with HL questions on balanced equations, mole calculations, limiting reactants, yield and atom economy.
Phosphoryl chloride, POCl3, is a dehydrating agent.
State the balanced chemical equation for the reaction of PCl3(1) with water.
PCl3(l)+3H2O(l)→H3PO3(aq)+3HCl(aq);
Two groups of students (Group A and Group B) carried out a project* on the chemistry of some group 7 elements (the halogens) and their compounds.
In the first part of the project, the two groups had a sample of iodine monochloride (a corrosive brown liquid) prepared for them by their teacher using the following reaction.
The following data were recorded.
The iodine used in the reaction was in excess. Determine the theoretical yield, in g , of ICl(l).
n(Cl_2)=(2.24/(2 x 35.45)=) 0.0316 mol;
n(ICl)=(2 x 0.0316)=0.0632 mol;
m(ICl)=(0.0632 x 162.35)=10.3 g;
Calculate the percentage yield of ICl(l).
(10.38.60×100=)83.5%;
The students reacted ICl(l) with CsBr(s) to form a yellow solid, CsICl2( s), as one of the products. CsICl2( s) has been found to produce very pure CsCl(s) which is used in cancer treatment.
To confirm the composition of the yellow solid, Group A determined the amount of iodine in 0.2015 g of CsICl2( s) by titrating it with 0.0500moldm−3Na2 S2O3(aq). The following data were recorded for the titration.
The overall reaction taking place during the titration is:
Calculate the amount, in mol, of iodine atoms, I , present in the sample of CsICl2( s).
(0.5×1.21×10−3)=6.05×10−4/0.000605( mol);
In this project the students explored several aspects of the chemistry of the halogens. In the original preparation of ICl(1), they observed the yellow-green colour of chlorine gas, Cl2( g), reacting with solid iodine, I2( s).
Chlorine can also react with water. State the balanced chemical equation for the reaction of Cl2( g) with water.
Cl2(aq)+H2O(l)⇌HCl(aq)+HOCl(aq);
Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula PH3.
2.478 g of white phosphorus was used to make phosphine according to the equation:
This phosphorus was reacted with 100.0 cm3 of 5.00 moldm−3 aqueous sodium hydroxide. Deduce, showing your working, which was the limiting reagent.
n(NaOH)= 0.1000 x 5.00=-0.500 «mol» AND P_4 /phosphorus is limiting reagent
Marking guidance:
Accept n(H_2 O)=100/18=5.50 A N D P_4 is limiting reagent.
Determine the excess amount, in mol , of the other reagent.
amount in excess <=0.500−(3×0.02000)00=0.440 «mol»
Impurities cause phosphine to ignite spontaneously in air to form an oxide of phosphorus and water.
State the equation for the reaction of this oxide of phosphorus with water.
P_4 O_10( s)+6 H_2 O(l) 4 H_3 PO_4(aq)
Marking guidance:
Accept P_4 O_10( s)+2 H_2 O(l) 4 HPO_3(aq) (initial reaction)
Accept P_2 O_5( s)+3 H_2 O(l) 2 H_3 PO_4(aq)
Accept equations for P_4 O_6 / P_2 O_3 if given in d (iii).
Accept any ionized form of the acids as the products.