Lithium chlorate is decomposed with heat to give lithium chloride and oxygen gas

Question

Lithium chlorate is decomposed with heat to give lithium chloride and oxygen gas. If 1.115 g of lithium chlorate is decomposed, how many milliliters of oxygen gas is released at STP? 2 LiCIO_3 (s) rightarrow^delta 2 LiCI(s) + 3 O_2 (g) Lithium metal reads with water to give lithium hydroxide and hydrogen gas. If 75.5 mL of hydrogen gas is produced at STP, what is the mass of lithium metal that reacted? 2 Li(s) + 2 H_2 O (l) rightarrow 2 LiOH(aq) + H_2 (g) (optional) The Solvay process is used to manufacture baking soda, NaHCO_3. In the process, CO_2, NH_3, H_2 O, and NaCI react to produce baking soda If 25.0 L CO_2 and 20.0 L NH_3 react at STP, with excess water and sodium chloride, what is the limiting reactant? Calculate the mass of baking soda produced. CO_2 (g) + NH_3 (g) + H_2 O (l) + NaCl (s) rightarrow NaHCO_3 (s) + NH_4 Cl (aq)

Explanation / Answer

3. molar mass of LiClO3 = 106.4 g/mol

moles of LiClO3 = 1.115 / 106.4 = 0.0105 mol

2 moles of   LiClO3 produces 3 moles of O2

1 moles of   LiClO3 produces = 3/2 = 1.5 moles of O2

0.0105 mol of   LiClO3 produces = 1.5 x 0.0105 moles of O2

= 0.0158 moles of O2

at STP, 1 mol O2 = 22.4 L O2

0.0158 mol O2 = 22.4 x 0.0158 L

= 22.4 x 0.0158 x 1000 mL = 354 mL

4. number of moles in 75.5 mL H2 = 75.5 / (22.4 x 1000) = 3.37 mmol

Amount of Li required to prodcuce1 mol H2 = 2 mol of Li

Amount of Li required to prodcuce 1 mmol H2 = 2 mmol of Li

Amount of Li required to prodcuce 3.37 mmol H2 = 2 x 3.37 mmol of Li = 6.74 mmol

atomic mass of Li = 7 g/mol

amount of Li required = 6.74 x 7 mg = 47.2 mg

5. number of moles of CO2 = 25 L / 22.4 L/mol = 1.116 mol

number of moles of NH3 = 20 L/22/4 L/mol = 0.89 mol

1 mol CO2 will react with 1 mol of NH3. So ammonia is the limiting reagent.

Molar mass of baking soda, NaHCO3 = 84 g/mol

mass of baking soda produced = 0.89 x 84 = 74.76 g

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