Question

WOL 2. What volume of 0.250 M HCl is required to completely react with 0.350 g of Na2CO3? Na2CO3(s) + 2HCl(aq) + 2NaCl(aq) + H2O(l) + CO2(g) 3. Calcium carbonate decomposes at high temperatures to give calcium oxide carbon dioxide. What volume of CO2 will be collected at 950 torr and 500°C when 75.0 g of CaCO3 react? CaCO3(s) → Cao(s) + CO2(g) 4. How many mL of 0.250 M KMnO4 are needed to react with 3.36 g of FeSO4 10 FeSO4 + 2 KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + 8H2O + K2SO 5. Potassium chlorate reacts with hydrochloric acid to produce chlorine gas. What mass of KCIO3 is needed to make 4.50 L of chlorine at 2.35 atm and 37.0°C? KCIO3(g) + 6HCl(aq) → KCl(aq) + 3Cl2(g) + 3H2O(1)

Answer 1

3 Na Coz (S) + 2 Hel cag) znace cag) +thore) & cozig) above eqh is balanced I mole of Naz coz (s) grequired 2 mole of tice even Ma, CO3 = 0.350 g i-e no of mole of Ma, cos = 0.0033 mole 0.00 33 mole ela, coz require 0.0066 mole of Hce. tence. no. of mole of nice = 0.250 x v. V= 0.0066 = 0.0264 L = 26.4 15ml 0.250 tence we require 26.415 me Hel (0.250 M) to react with 0.350g of Na2CO3 100 (3) CaCO3(s) - 0 (s) & Co (8) - I mole of caco give Imole of coz. geven. :- 75g of call3 = 750-75 mole of cdo P = 950 rose Ta soo'c = 773k 0 475 mole of CaCO3 give 0.75 mole of CO2 (8) So. PV = nRT [R = 62.3637 L-Tees (mol. K7 V = nRT = 0.75 mblx 62.36372 Tore srl xx 773X 950 tons 75g of caloz ve 38.058 2 of colg). P V= 38.058L

A lofesoy + 2 KMnO4 + 8 H₂SO45 fe, (SO4)3 + 2 MnSO4 + 8H20 + K₂504 from abave equ To mole of resou react with 2 mole of KMnO4" I mole of resou euad with 2 mol of KMnO4. 12 I liven 3.369 resou meles of FeSO4-336 0.0221 moces 152 Hince 0.0221 mole of resou react with 3 xo.02 -0.0042 meleg KMnOy. Volume of kmn Oy = 0.0042 0-250 = 0.0168L = 16.8 me [Volume of knnoy = 16.8 me) Hence 8.36g of resoy required 16.8 me of 0.250m k mnoy. (5) KOOz (g) + 6nce caq) → Kecaq) + 39 (g) + 3 Hold) - Imece of keloz is needed to produce 3 mole of chig) given : moles of clz = (m) PV = nRT m = PV = 2.35 atm x 4. sox RT 0.08214-ohm/mak 310 K 1o 575 mole 041SS rol 25.us

e of moles of Cl2 given = 0.4155 mole 3 moll Cl, is produced by Imole kloa. I more ce, it produced by / mole of KClOz 0.4155 mole clz is produced by 0-4155 = 0.138 mole of koz # mo. moles o KUOz - 0.138 Civen mais no. of moles: Molar mass = 122.6 Molar mass of keeoz mass of kl 03 (reacted) = 0.138x 122.6 ce = 16.989 Hence 16.98g KClOz is meoded to make 4.50 L og 9.35 atm and 37°c