Question

For each of the following unbalanced equations, calculate how many grams of each product would be produced by complete reaction of 14.2 g of the first reactant. (a) CO2(g) + H2(g) → CH4(9) + H20(1) CH4 4.9.) 14.2 x g H20 49 11.618 X Check the number of significant figures. g (b) TiBr4(9) + H2(g) → Ti(s) + HBr(9) Ti 4.9.169.9 X 9 HBr 4.9) 1011.37 x g (c) NaOH(s) + CO2(g) → Na2CO3(s) + H20(1) Na2CO3 407 H20 49 (d) NO(g) + H2(9) ► N2(9) + H20(1) N2 49 H2O 49

Answer 1

I Pogo Mi Salution First we have balanced all the reaction. (6) CO2(g) + tholg) - CHA (g) + H20(e) First balance the Oxygen (0) con (G) + H2(g) - CH4 (g) + 2H2O(g) Second balance the hydrogen CH) I CO2(g) + 4H2(g) - CHA(g) + 2H2(g) In balance equation 1 mole of first reactant (102) produced 1 mole of CH4 Dand 2 male Of H20 molecular mass of Co2 = 44g Imol So I mole of co2 = 44g. 44g of con produced 1 male of CH4 and a mole - Of H₂O Molecular mass of CH4 = 16 g/mol Molecular mass of Imol H2O = Imole of CH4 = 16g 2 mole of H2O= 2x10 = 36g.

dag of Co, pтоди са 2 - 18 од она 14 top priдасад = 16 а у сня — J4.2 Я оf со 2 рто дисә -16 14.2 ) 44 Б.16 ОГСН4 Б.16 2 of o4 44 g оf со родалее – 562 оf нэр Тя оf Co2 pxoдисер - 36 ) a of 12.0 - 14-24 of copvo дисед = 56 14.2 ) = 11-62 8 of H20 CH4 – H2O = Б.16 1.629 T'B«4 (9) + H» (Я) - T'S) + HBr (9) Balance the first balance reaction al the Bor in both direction. TiB4 + H T5 + 4HBs second balance the H in bath direction.

| Ti B4 () + 2H2 (9) » Т(с) + 4HBя 1 mole of Ti'Вт4 viдисед 1 mole of T" ол | 4 mole of Hre i mole of Ti Br4 = molecular mass of — 364 Бяров Tigra 1 mole of T = 42. Сая 4 mole Of HBr = 4x molecular mass of H Br. — 4x (Go •91) а — 323.64 . З 69.5 g of Ti ere produce = 45» ва от sa, 11 of Ti B*4 ходисе = 47.&огт, 362.5) S6, 14:24 { ту ва роди се = ( 49.87 , 14,2 ) а 4. В 14-2 ) = 1.85 от Again 3 _369.5 4 оf TTB+4 родые 2 Заз. 64g of HB 19 of TiB-4 рябие - 323.64 o of HB 36. s of HB,

So, 14.2 g of Tibro produce =/323.64 x 14.2g 364.5 = 12.5 g of HBr. Ti = 1.85 g HBr = 12,5 g AOC NaOH (s) + Coag - Na2CO3(s) + H2ole) Balance the 2NaOH (s) + 0 in both CO2(g) side. Narco, (s) + H20(e) of Here 2 male of NaOH Na2CO3 and 1 mole produced 1 mole of H2O. 1 2 male of NaoH = 2x molecular mass of Nach = 24 409 12 male of Nach = Bog I mole of Na2CO3 = molecular mass of Na2CO3 = 106 g I mole of H2O = 18g

Pase No SO Во я вон рубда се = Jo6 я оf аэоз тя Noон родите - 106 я оf a 2 CO 3 14. 24 часH produce = 106 14, 22 Со і 28 18. 62 оf нар боз again Во я fаон род се = 10 of H2 о 1я Наон „ртори се = 18 я оf B2,0 14- 24 Гоо + ртодом се = (18 x 14.2 ) = 3. 2 Я of H20 _fa 2 CO3 = 18. В Нэр = 3, 21° | Pre" 2. , а о т Ang. Nog) + H₂(g) + Ng Cg 1 + H2ole) First balance the ff in both side N2 + H2O 2 No + H2 Second balance the o in both side

2rot H2 - N2 + 2H2O Then balance the H in both side 200 g) + 2H2G) - 1289) + 2 H20les Here 2 mole of produced i male H2O, No (first reactant) of NoD and a mole of We know that - 2 mole of No = 2x molecular mass of No = (x 30 g = 60g I mole of Na= 20 g 2 male of H2O = 36g. 1 Here 60g o produce = 20g of fra 1 19 No produce CO 14.29 No produece = (26 x 14,2 19 = 6.63g Of N2

again, 6оя у го одисе = 36 Я оf H2.0 19 ар No рrода са = 36 а од но 14.28 оf о рода се = (36 14. 13 = 0-5 2 я у нэр N2 = 6.63 Я. нэ он S52 а