Question

need help on d, e and f

957 77 9 C 3. Cyanogen, C2N2 is a colorless, poisonous gas that can be prepared by the following: 2 CuSO4 (a)+4 KCN(a)CN2 +2 K2SO4 (a) + 2 CuCN () a Determine how many moles of KCN are required to react with 3.7 moles of CuSO 3.7 mal CusO ml CuSO ml Cy tml Cusa 2mol Cus 7.4moles Ka b. Determine how many moles of cyanogen are produced from 1.75 moles of CuSO mol Cala 2 mol Cuso4 175moles CuSO c. Determine how many moles of K2SO4 are produced from 5.25 moles of CuSO4 5.25 moles cusoxmal KaS0 5.25 mbles Ks0 mot CoS Determine how many grams of KCN are required to react with 2.5 moles of CuSO4 2.5 males Cu So Tmol CuSl Determine how many grams of cyanogen are produced from 3.5 moles of CuSO4 e. Determine how many grams of cyanogen are produced from 45.3 g of KCN f.

Answer 1

a. From the equation of the reaction: 2 moles of CuSO₄ need 4 moles of KCN. 3.7 moles will require: 4 / 2 * 3.7 = 7.4 moles of KCN

b. Moles of cyanogen = moles of CuSO₄ * 1 / 2 = 1.75 / 2 = 0.875 moles

c. Moles of K₂SO₄ formed = moles of CuSO₄ reacted = 5.25 moles

d. Moles of KCN = 2 * moles of CuSO₄ = 2 * 2.5 = 5 moles

Mass = moles * molar mass = 5 * (39 + 12 + 14) = 325 g

e. Moles of cyanogen = 0.5 * moles of CuSO₄ = 0.5 * 3.5 = 1.75 moles

Mass = moles * molar mass = 1.75 * (2 * 12 + 2 * 14) = 91 g

f. Moles of KCN = mass of KCN / molar mass = 45.3 / 65 = 0.697 moles

Moles of cyanogen = 1 / 4 * moles of KCN = 1/ 4 * 0.697 = 0.174 moles

Mass of cyanogen = moles * molar mass = 0.174 * 42 = 7.308 g