At STP the volume of HCI that will result from the reaction of 0.680 L of H_2 an

Question

At STP the volume of HCI that will result from the reaction of 0.680 L of H_2 and 0.750 L of CI_2 is ___. H_2(g) + CI_2(g) rightarrow 2 HCI(g) A) 0.680 L B) 0.750 L C) 1.36 L D) 0.340 L E) 1.50 L Commercial ammonia is produced by reacting hydrogen gas with nitrogen gas according to the following equation. N_2(g) + 3 H_2(g) rightarrow 2NH_3(g) If 6.00 times of N_2 are reacted with 1.2 times 10^5 L of H_2, the limiting reactant is ____ and the volume of ammonia produced is ____ L. A) N_2 3.33 times 10^3 B) H_2 6.00 times 1^3 C) N_2 8.00 times 10^4 D) H_2 1.8 times 10^4 E) H_2 8.00 times 10^4

Explanation / Answer

(6)

H2 (g) + Cl2 (g) -------------> 2 HCl (g)

According to above equation and According to Dalton's law of equal volumes,

1 volume of H2 can react with 1 volume of Cl2 to form 2 volumes of HCl.

Based on the given data, H2 is limiting reagent,

So, 1 volume of H2 can form 2 volume of HCl

then, 0.680 L of H2 can form 2 * 0.680 = 1.36 L of HCl

So, the answer is (C)

(7)

N2 (g) + 3 H2 (g) ------------> 2 NH3 (g)

According to the above balanced equation,

1 volume of N2 = 3 volumes of H2

then, 6.00 * 104 L of N2 requires 3 * 6.00 * 104 = 1.8 * 105 L of H2 ( > 1.2 * 105)

So, H2 is limiting reagent.

3 volume of H2 can produce 2 volumes of NH3

then, 1.20 * 105 L of H2 can produce 2 * 1.20 * 105 / 3 L = 8.00 * 104 L of NH3

So, the answer is (E)

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