How many liters of oxygen gas at 153 °C and 0.820 atm can be prepared by the decomposition of 22.4 g of solid KCIO3?
2 KCIO3(s) > 2 KCl(s) + 3 O2(g)

V =  11.51 L

Explanation:

Given data:

Volume of oxygen gas produced = ?

Pressure = 0.820 atm

Temperature = 153°C (153+273 = 426K)

Mass of KClO₃ = 22.4 g

Solution:

Chemical equation:

2KClO₃    →         2KCl  + 3 O₂

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles =  22.4 g/  122.5 g/mol

Number of moles = 0.18 mol

now we will compare the moles of KClO₃  and O₂

                       KClO₃           :              O₂

                           2               :               3

                         0.18             :             3/2×0.18 = 0.27

Volume of oxygen:

PV = nRT

R = general gas constant (0.0821 atm.L/mol.K)

0.820 atm× V = 0.27 mol × 0.0821 atm.L/mol.K ×426 K

V = 9.44 atm.L / 0.820 atm

V =  11.51 L

the answer is b because, investigating whether another species of moth also drinks tears might be interesting, but it would not directly explain why this species of moth does.

to explain why this species of moth drinks tears, the scientist could ask: "what nutrients that moths need are in elephant tears? "

the scientist could experiment with moth repellent or oter ways to keep the moths away, but that would not answer his original question.