Question

Given the unbalanced chemical equation, Cshe c!+0,(9) + COCg+H, OCI) which of the following correctly expresses the mole ratio between co and Oxygen al 5 mole CO /41 mole da b/ 10 mole co/41 mole og ci Imole coli mole og dl 10 mole C0/6.5 mole 0, Given the chemical equation, Zn (s) + 2HCI Caq) → ZnCl, Cag) + H, (g), How many molecules of He will be produced og 11 atoms Zn are consumed in the reaction? Given the balanced equation, Na, SA Cag) + BaCl, (aq) → Nacicaq) +BaSO4(5), how many moles og BaCl, are required to produce 13499 Naciz Given the unbalance equaton, Cr(NO3), caq) + LiOH (aq)LiNog Cag). CrcoH), (s), how many grams og Cr Cou), are produced from 1.4699 LiOH ? 5 Given the unbalanced equation, NHH NO, (s)→ Ng(9)+ HOCI), how many grams of H, O are produced from the de composition or - 0.97 g of NHA NO?

Answer 1

① Unbalanced equation is :- CH2 (4) + O2 (g) - CO2 igo et thow After Balancing, the equation is o Cathe (1) + 802 g 6H₂011) + 50, 198 - from the unbalanced equation we can see that I mole Or produces I mole Logg in the medle to male ratio is 1:1 so the correct answer is option (6)

- Znco) + 240lags & Zn 2 lag of the igs from the reaction, we can see - I mole of en produces 1 nede the we know, I mole zn = 6.0224 62s atoms. & I moll the 2x 6.022 x 1023 atoms - 2 12.049_X003 alones. 27th 2n . 6022 x 10 23 ataus produces 12.044 x 10's atoms of the 2) 1 atoms of Zn produces ! = 12.044 x 105 x 11 aroms of the - 23 6*02281 023 2 2x11 atoms of H₂ е 22 осми 8 , As I Hydrogen (th) moleulle has 2 atoms of H. or, a atoms of 11 ls equal & mole – ule of M₂. 2 22 atodes of H & equal to 22 more - ule of the & 22 atoms of E 11 mole aile of the o ft 11 atours of En are consunced then 11 noleules of the is produced ' 2

③ Naz sa cag) + Bally cag) - 2 Nad cag, et Basoq is) from reaction we can see? 2 moles of Nall is produced from I mole Ball, 2 2 noles Wall = 2 (23 + 35.5 2 1179 Nada e I mole Ball, a (137.32 + 35.5x2g 2 208.32 g Ball 8 117g Nach produced by 208 32 g 2 1.34 g Nall produced by 208.32x1234 17 g of Balla 2.385 g Bally Now, 208.32 g Ballz & Imole Ball a) 2.3859 Ball ? t x 2385 mole 208.32 Bally 2 0.0114 mole Ball

) ti (0)2 сая) + L°00 to ) - L°/03 104) + Cocos кі вади? ° - Is) Cy (wg), + 2 оң - tr(о), * 34No, САр де Сафом ", 2 yлole L? ви е ходиш | ok Сяби), 2 моlе ери 3 1 + 16+ ( )c L°0И : 724 Рои о I wole (он), - ( + 16х3 + 3x) 4 Суби), (033 х (он), 0 5. 12, LvИ Aь-Алш 103 з су (он) 4. е) 1.46 3 L'on pльдаа цо3 x 1-4, G (он) - 2 - 08 Gя (он), 12 - Сх ( ои) , , The unbalanced equation is ! NHA NO₂ is N2 (g) + the (1) After Balancing ? Миа 02 (1) — М. 4) + 2 р )

from the given balanced equation, we get I mole NHA No produces 2 mole the I male NH Noa = (14x2 + 4x 1 4 16 x2) g. NHG NO2 с (2g+A+ за) 4 ли N0, 64 1 1A нь, 2 du Ho - 2 ( 2к ( + Tь) 44,0 2 дк 184 И, о o 26 о“ И, о во 643 ми, но, pkadar4 % по Ә0-174 м, 10, poda 6 x 6-12 а 4,0 = 0. 545 4 цо Thus, 0.97 g NHG NO2 produces 0- CAS" И,0 .