Formation of polyethene from calcium carbide takes place as follows : CaC 2 + H 2 O → Ca ( OH ) 2 + C 2 H 2 ; C 2 H 2 + H 2 → C 2 H 4 ; n C 2 H 4 → - CH 2 - CH 2 - n Determine the amount of polyethylene possibly obtainable from 64 . 0 kg CaC 2 can be:

The balanced chemical equations of process are as follows, CaC 2 + 2 H 2 O → Ca ( OH ) 2 + C 2 H 2 C 2 H 2 + H 2 → C 2 H 4 ; n C 2 H 4 → - CH 2 - CH 2 - n From the balanced chemical equations: mol of CaC 2 = mol of C 2 H 2 = mol of C 2 H 4 Moles of C 2 H 4 = mass of CaC 2 molar mass of CaC 2 = 64 × 10 3 64 = 10 3 mol Moles of polyethene = mol of C 2 H 4 n mass of polyethene molar mass of polyethene = mol of C 2 H 4 n mass of polyethene 28 n = 10 3 n Mass of polyethene = 28 × 10 3 g = 28 kg

E M B I B E

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 42

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

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Acetylene & butene have empirical formula $CH$ & ${CH}_{2}$ respectively. The molecular mass of acetylene and butene are $26$ & $56$ respectively deduce their molecular formula.

Important Questions on Mole Concept - 1

EASY

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 43

An oxide of nitrogen gave the following percentage composition :

N = 25.94

and

O = 74.06

. Calculate the empirical formula of the compound.

EASY

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IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 44

Find the density of

{CO}_{2} (g)

with respect to

N_{2} O (g)

.

MEDIUM

KVPY Aptitude Test - Stream SA

IMPORTANT

Practice Book for KVPY Aptitude Test - Stream SA Chemistry>Chapter 1 - Mole Concept - 1>EXERCISE>Q 45

Formation of polyethene from calcium carbide takes place as follows :
${CaC}_{2} + H_{2} O \to Ca (OH)_{2} + C_{2} H_{2};$ $C_{2} H_{2} + H_{2} \to C_{2} H_{4};$