Chemistry College

How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degrees Celsius and 1.02 atmospheres?

CS2(l) + 3O2(g) yields CO2(g) + 2SO2(g)


2.78 liters
5.95 liters
12.1 liters
11.9 liters

Answers

Answer 1

The volume of carbon dioxide produced is approximately (d) 11.9 liters.

To determine the amount of carbon dioxide (C[tex]O_2[/tex]) produced when 37.8 grams of carbon disulfide (C[tex]S_2[/tex]) reacts with excess oxygen gas ([tex]O_2[/tex]), we need to use stoichiometry and the given balanced chemical equation:

C[tex]S_2[/tex](l) + 3[tex]O_2[/tex](g) → C[tex]O_2[/tex](g) + 2S[tex]O_2[/tex](g)

First, we calculate the number of moles of C[tex]S_2[/tex] using its molar mass:

Molar mass of (C[tex]S_2[/tex]) = 12.01 g/mol (C) + 32.07 g/mol (S) × 2 = 76.14 g/mol

Number of moles of (C[tex]S_2[/tex]) = mass / molar mass = 37.8 g / 76.14 g/mol ≈ 0.496 mol

From the balanced equation, we can see that the stoichiometric ratio between (C[tex]S_2[/tex]) and C[tex]O_2[/tex] is 1:1. Therefore, the number of moles of C[tex]O_2[/tex] produced will also be 0.496 mol.

Now we can use the ideal gas law to calculate the volume of C[tex]O_2[/tex] at the given temperature and pressure. The ideal gas law equation is:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Converting the temperature from Celsius to Kelvin:

T(K) = 28.85°C + 273.15 = 302 K

Using the ideal gas law:

V = nRT / P = (0.496 mol) × (0.0821 L·atm/mol·K) × (302 K) / (1.02 atm) ≈ 11.9 L

The correct answer is 11.9 liters.

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Related Questions

HQ5.39
Homework Answered Due Today, 11:59 PM
Consider the reaction of hydrogen with sulfur to form hydrogen sulfide: 8H₂(g) + S8 (s)-8H₂S(g), which has an enthalpy of
reaction of -20.15 kJ/mol. If 101 g of sulfur are reacted with an excess of hydrogen, how much heat is produced (kJ)?

Answers

Answer: 63.598 kJ

Explanation: 1. calculate No.of mols

101 g/32 g/mol = 3.15625 mol

2. from the reaction 1 mol of sulfur give 20.15kJ

so 3.15625 mol will give : 3.15625*20.15 = 63.598 kJ

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