A sample of lead (II) cyanide contains 0.660 moles of nitrogen. What mass Pb(CN)2 is present?

Question

A) 85.5g Pb(CN)2
B) 428g Pb(CN)2
C) 342g Pb(CN)2
D)2.00g Pb(CN)2
E) 518g Pb(CN)2

Answer 1

Pb(CN)2 --> Pb(2+) + 2 C(4+) + 2 N(5-)



mole of Pb(CN)2 = 1/2 * mole of N [ratio of moles = ratio of coefficients]

= 1/2 * 0.660 = 0.330 mol



mass of Pb(CN)2 = 0.330 * 259 = 85.47 g = 85.5 g (A)

Answer 2

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RE:

A sample of lead (II) cyanide contains 0.660 moles of nitrogen. What mass Pb(CN)2 is present?

A) 85.5g Pb(CN)2

B) 428g Pb(CN)2

C) 342g Pb(CN)2

D)2.00g Pb(CN)2

E) 518g Pb(CN)2

Answer 3

Lead Ii Cyanide

Answer 4

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