6th question

6th question salt the following composition:• A1 10 K 1 S 24 the gen. Calculate the empirical formula of the saltv IAI 27, K — 39, S 32. O 161 i Ans. 2.3 CHEMICAL EQUATIONS 1. What mass of silver chloride will obtained bv adding an of hydrochloric acid to a solution of 0.34 g of silver nitrate. | .Ag=108, N—14, 0—16, I IAns. 0.287 gl 2 What volume of oxygen at s.t.p. will be Obtained by the action of heat on 20 g. of KCIOA 1 K-39, | IAns. 5.486 It,.l 3. From the equation : 3Cu + 8HN03 + 4H20 + 2NO. Calculate (i) the mass of copper needed to react with 63 g of nitric acid (Ans. 24 (ii) the volume of nitric oxide collected at the same time. [Cu=64, Hal, 0=16, N=141tAns.5.6 4. Zinc blende [ZnSJ is roasted in air. Calculate : al the number of moles of sulphur dioxide liberated by 776 g of ZnS and (Ans. 8 molesl b] The weight of ZnSrequired to produce 22.4 lits Of S02 at s.t.p. IS=32, Zn=65, 0=161 (Ans. 97 g. I Ammonia reacts with sulphuric acid to give the fertilizer ammonium sulphate. Calculate the volume of ammonia [at s.t.p.] used to form 59 g of ammonium sulphate. [Ans. 20.02 Its.l [ N=14, H=l, S=32, 0=161. Heat on lead nitrate gives yellow lead [II] oxide, nitrogen dioxide & oxygen. Calculate the total volume of N02 & 02 produced on heating 8.5 of lead nitrate. [P b = 207, N = 14, O = 161. [Ans. 1.15 of N02 & 0.287 of 02 (1.437 Its.)l 2KC103 2KCl + 302; C + 02 C02. Calculate the amount of KC103 which on thermal decomposition gives 'X' vol. of 02, which is the volume required for combustion of 24 g. of carbon. [Ans. 163.33 g.l (K = 39, a = 35.5, 0=16, c = 121. :alculate the weight of ammonia gas. Required for reacting with sulphuric acid to give 78 g. of fertilizer ammonium sulphate.

Dear student
Please find below the solution to the asked query:
2Pb(NO₃)₂ (s) → 2PbO(s) +4NO₂ (g) + O₂ (g)

Molar mass of Pb(NO₃)₂ = 331 g/mol

Molar mass of NO₂ = 46 g/mol

Molar mass of O₂ = 32 g/mol

Now, we know that molar mass of a gas contains 22.4 L of gas so,

46 g of Nitrogen dioxide = 22.4 L

32 g of Oxygen = 22.4 L

According to balanced chemical equation,

(2x331)=662g of lead nitrate yields (4x22.4)=89.6 L of nitrogen dioxide and 22.4 L of oxygen gas.

Total volume of gases = 89.6 + 22.4 = 112 L

662 g of lead nitrate produces 112 L of gases

8.5 g of lead nitrate will produce (112x8.5)/662 = 1.44 L of gases

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