How many moles of ammonia can be made by reacting 12 and of N_2 with 17 mol of H

Question

How many moles of ammonia can be made by reacting 12 and of N_2 with 17 mol of H_2? N_2(g) + 3 H_2(g) rightarrow NH_3(g) a. 24 mol NH_3 b. 12 mol NH_3 c. 17 mol NH_3 d. 11 mol NH_3 e. 29 mol NH_3 How many molecules of O_2 would react with 56 C_2H_6 molecules according to the following balanced equation? 2 C_2H_6 + 9 O_2 rightarrow 4 CO_2 + 6 H_2O a. 784 molecules O_2 b. 329 molecules O_2 c. 196 molecules O_2 d. 112 molecules O_2 e. 50 molecules O_2 What type of reaction is listed below 2C_2H_6 + 7O_2 rightarrow 4CO_2 + 6H_2O a. Combustion reaction b. Acid Based reaction c. Decomposition reaction d. Combustion reaction e. Precipitation reaction

Explanation / Answer

(11)

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

From above equation, 1 mol of N2 requires 3 mol of H2

then, 12 mol of N2 requires 36 mol of H2.

But there is only 17 mol H2. SO, H2 is limiting reagent.

And according to the balanced equation 3 mol H2 can from 2 mol NH3

Then,

17 mol H2 can form 17 * 2 / 3 = 11.33 mol NH3

So andwer is (d)

(12)

2 C2H6 (g) + 7 O2 (g) ---------------> 4 CO2 (g) + 6 H2O (g)

From the above balanced equation,

2 molecuels of C2H6 react with 7 molecules of O2

Then, 56 molecules of C2H6 react with 56 * 7 / 2 = 196 molecuels of O2

So the answer is (c)

(13)

(d) Combustion reaction

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