Question

For the reaction: 3 H_2 (g) + N_2 rightarrow 2 NH_3 (g) 3mol H_2 is reacted with 6 mol N_2 mol of NH_3 is produced mol H_2 remains mol N_2 remains For the reaction 2 N_2H_4 (l) + N_2O_4 (l) rightarrow 3 N_2 (g) + 4 H_2O (l) 160 g N_2H_4 is mixed with 160 g N_2O_4 is the limiting reagent g H_2O is produced For the reaction Fe_2O_3 (s) + 3 CO (g) rightarrow 2 Fe (g) + 3 CO_2 224 g of CO is available to react with 400 g Fe_2O_3 is the limiting reagent g of iron is produced g of CO_2 is produced For the reaction 2 C4H10 (g) + 13 O2 (g) rightarrow 8 CO2 (g) + 10 H2O (l) 300 g of C4H10 is combusted in 1000 g of O2 is the limiting reagent g H_2O is produced

Might you show me step by step in order to undestand everything.

Thanks a lot in advance.

Answer 1

1. The balanced equation is N₂ + 3 H₂ = 2 NH₃

From this, it is clear that 3 moles H₂ can react with a maximum of 1 mol N₂ which produces 2 mol NH₃.

So, even if 6 moles N₂ is present, 3 moles H₂ can react with only 1 mol N₂.

Hence,

(a) 2 mol NH₃ is produced

(b) 0 mol H₂ remains

(c) 5 mol N₂ remains

2. Here, 1 mol N₂O₄ reacts with 2 mol N₂H₄.

Now, 160 g N₂O₄ means 1.75 moles (since its molecular wt. = 92).

So, 1.75 mol N₂O₄ reacts with 3.5 mol N₂H₄.

3.5 mol N₂H₄ means 112 g N₂H₄

(a) Limiting reagent is N₂O₄ as it gets exhausted completely.

(b) For 1 mol N₂O₄ , 4 mol H₂O is produced.

So, for 1.75 mol N₂O₄ , 7 mol H₂O is produced.

i.e. 126 g H₂O is produced (M.W. = 18).

3. Here, 1 mol Fe₂O₃ reacts with 3 mol CO.

400 g Fe₂O₃ means 2.5 moles Fe₂O₃

So, 2.5 moles Fe₂O₃ reacts with 7.5 mol CO.

7.5 mol CO = 210 g CO (MW = 28).

(a) Limiting reagent is Fe₂O₃

(b) For 1 mol Fe₂O₃ moles of Fe produced = 2 mol

So, for 2.5 mol Fe₂O₃ moles of Fe produced = 5 mol = 280 g of Fe (atomic wt of Fe = 55.85)

(c) For 1 mol Fe₂O₃ moles of CO₂ produced = 3 mol

So, for 2.5 mol Fe₂O₃ moles of CO₂ produced = 7.5 = 330 g of CO₂ (Molc. wt of CO₂ = 44)

4. Here, 1000 g O₂ = 31.25 mol O₂

Since 13 mol O₂ reacts with 2 mol C₄H₁₀ so 31.25 mol reacts with 4.8 mol C₄H₁₀

i.e. 280 g C₄H₁₀

(a) Hence, O₂ is the limiting reagent.

(b) 24 moles H₂O is produced from 31.2 mol O₂.

i.e. 432 g H₂O is produced.