5. A compound contains 40.0% C, 6.71% H, and 53.29% O by mass. The molecular wei

Question

5. A compound contains 40.0% C, 6.71% H, and 53.29% O by mass. The molecular weight of the compound is 60.05 amu. The molecular formula of this compound is- A) C2H402 B) CH20 C) C2H304 D) C2H204 E) CHO2 6. The combustion of propane (C3Hs) in the presence of excess oxygen yields Co2 and H20: C3H8 (g) + 502(g) 3CO2 (g) + 4H20 (g) when 2.5 mol of O2 are consumed in their reaction,-- A) 1.5 B) 3.0 C) 4.2 D) 7.5 E) 2.5 mol of CO2 are produced. - Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia 7. N2(g) + 3H2 (g) 2NH3 (g) Determine the mass of N2 (g) required to completely react with 18.5 g of H2 A) 0.51 B) 771 C) 257 D) 85.7 E) 6.17 8. The combustion of ammonia in the presence of oxygen yields NO2 and H20 4NH3 (g) + 702 (g) 4NO2 (g) + 6H20 (g) The combustion of 43.9 g of ammonia with 258 g of oxygen produces A) 212 B) 178 C) 119 D) 0.954 E) 43.9 g of NO2.

Explanation / Answer

Answer 9:

Mole ratio of C: H : O = 40.0 /12 : 6.71 /1 : 53.29/16 (deviding by atomic number)

= 3.333 : 6.71 : 3.33

= 1 : 2 : 1 (deviding by the smallest number)

So empirical formula is (CH2O)n and MW is 60.05

So, (12 + 2 x 2 + 16 ) x n = 60.05

then, n =2

So, molecular formula is  (CH2O)n i.e  (CH2O)2  or C2H4O2   (Option A is the correct answer).

Answer 6.

5 mole of O2 is producing 3 mole of CO2

So, 2.5 mole of O2 will produce ( 3 x 2.5 ) / 5 = 1.5 mole of CO2 (Option A is the correct answer).

Answer 7.

1 mole of N2 is reacting with 3 mole of H2

Now MW of N2 is 28.01 and H2 is 2.016

So, ( 3 x 2.016 ) = 6.048 g of H2 will react with 28.01 g of N2

Then , 18.5 g will H2 will react with (28.01 x 18.5 ) /6.048 = 85.67 g of N2 (Option D is the correct answer).

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