What volume of carbon dioxide is produced when 6.40 g of methane gas, CH4 (g), reacts with excess oxygen? All gases are at 35.0C and 100.0 kPa.

V = 10.3 L

Explanation:

Given data:

Mass of methane = 6.40 g

Volume of CO₂ produced = ?

Temperature = 35°C (35+273 = 308 K)

Pressure = 100.0 KPa (100.0/101 = 0.98 atm)

Solution:

Chemical equation:

CH₄ +  2O₂        →       CO₂ + 2H₂O

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 6.40 g/ 16 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of CO₂ with CH₄.

         CH₄          :         CO₂

           1             :           1

        0.4            :         0.4

Volume of CO₂:

Formula:

PV = nRT

0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K ×  308 K

0.98 atm ×V = 10.11 atm.L

V = 10.11 atm.L /0.98 atm

V = 10.3 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.