Pearson Edexcel IAL Chemistry 4.3 Reducing practical risks

Practise choosing practical precautions that reduce chemical risk, using the reagent, hazard label and procedure step given.

Exam points

explain why a step needs ventilation when toxic vapours or gases may be produced
use tables of state and hazard symbols to justify safer handling of one reagent over another

[Maximum number: 2]

This question is about ethene, $\mathrm{C}_{2} \mathrm{H}_{4}$ .

Ethene and chlorine react to form 1,2-dichloroethane.

One hazard symbol for 1,2-dichloroethane is shown.

Figure for Question 4.3—Suggest ways in which risks can be reduced and reactions carried out safely, for example: i working on a smaller scale ii question 1 — Edexcel A-Level Chemistry AS

By identifying this hazard, give one way of minimising the risk when working with 1,2-dichloroethane in a laboratory.

Assume the use of safety goggles and a laboratory coat.

(d)(i) Method 1 - calculation of mass C, H and O (1) - calculation of moles C, H and O and Ignore $\ce{SF}$ except 1 $\ce{SF}$ for all methods Example of calculation: mass $\ce{C=24}$ /28 x 10.0=8.5714( g) mass $\ce{H=4}$ /28 x 10.0=1.4286( g) mass $\ce{O=15.7-}$ 10.0=5.70( g)arrayc:c:c C &: & H 8.5714/12 &: & 1.4286/1 array: $\ce{O5.7}$ empirical formula empirical formula is C_2 H_4 O Method 2 - calculation of moles C_2 H_4 and O 1 calculation of moles C_2 H_4 and O_2 (moles C_2 H_4=10.0/28= ) 0.35714 (. moles .O_2=(15.7-10.0)/32=) 0.17813 - empirical formula M2 dependent on M1 empirical formula is C_2 H_4 O Method 3 - calculation of moles C_2 H_4 and (moles .C_2 H_4=) 10.0/28=0.35714 (2) M_r product (1) (M_r. product .=) 15.7/0.35714=43.96 - empirical formula and calculation of M_r of empirical formula (1) M2 dependent on M1 empirical formula is C_2 H_4 $\ce{O2}$ x 12+4 x 1+1 x 16=44

All question bank results loaded

4.3—Suggest ways in which risks can be reduced and reactions carried out safely, for example: i working on a smaller scale ii question 1