Question

1. how many grams of Pb are formed when 0.105 g NH3 react
2. If a reaction vessel contains 0.15 moles of LiOH and 0.08 moles CO2 which compound is the limiting reagent
3. Lithium hydroxide absorbs carbon dioxide forming lithium carbonate and water
2 LiOH + CO2 Li2CO3 + H 2 0
4.

Answer 1

36604, Inorganic Chemistry

1. how many grams of Pb are formed when 0.105 g NH₃ react

2. If a reaction vessel contains 0.15 moles of LiOH and 0.08 moles CO₂ which compound is the limiting reagent

3. Lithium hydroxide absorbs carbon dioxide forming lithium carbonate and water

$$2 \mathrm{LiOH} + \mathrm{CO_2} \mathrm{Li_2CO_3} + \mathrm{H_2O}$$

Solution:

1) Lead(II) oxide reacts with ammonia as follows:

$$3 \mathrm{PbO} (s) + 2 \mathrm{NH_3} (g) \longrightarrow 3 \mathrm{Pb} (s) + \mathrm{N_2} (g) + 3 \mathrm{H_2O}$$

The number of moles NH₃ in this reaction is: $n = \frac{0.105}{17} = 0.006$ moles.

In accordance with the reaction, if 2 moles of NH₃ react 3 moles of Pb will be formed.

So, if react $n = \frac{0.105}{17} = 0.006$ moles of NH₃ $\frac{0.006 \cdot 3}{2} = 0.009$ moles of Pb will be formed.

Therefore, the mass of Pb is 0.009·207.2=1.865 g.

**Answer: $m(\mathrm{Pb}) = 1.865\ \mathrm{g}$**

2) Lithium hydroxide reacts with carbon dioxide as follows:

$$2 \mathrm{LiOH} + \mathrm{CO_2} = \mathrm{Li_2CO_3} + \mathrm{H_2O}$$

So, 2 moles of LiOH react with 1 mole of CO₂, or 0.15 moles of LiOH react with $\frac{0.15 \cdot 1}{2} = 0.075$ moles of CO₂. Therefore, if a reaction vessel contains 0.15 moles of LiOH and 0.08 moles CO₂, LiOH is the limiting reagent.

**Answer: LiOH is the limiting reagent.**