What mass of water is produced from the complete combustion of 5.50×10−3g of methane?

1 answer:

3 0

The reaction equation is CH4 + 2O2 = CO2 + 2H2O. The ratio of mol number of water and methane is 1:2. The mol number of methane is 5.5*10^(-3)/16=3.4*10^(-4) mol. The mass of water is 18*3.4*10^(-4)*2=1.2*10^(-2) g.

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Identify the molecule containing the least oxygen atoms. A) 12.044 × 10^23 molecules of CO2. B) 12 g of O3. C) 0.1 mol of V2O5.

olga_2 [115]

Answer:

C. 0.1 mol of V₂O₅

Explanation:

Number of moles of oxygen atoms present in each molecule is derived as follows:

A. Number of molecules of CO₂ in 12.044 * 10²³ molecules of CO₂ = 12.044 * 10²³/6.022 * 10²³ = 2 moles of CO₂ molecules

Each molecule of CO₂ contains 2 atoms of oxygen. Therefore, number of moles of oxygen atom present in 2 molecules of CO₂ = 2 * 2 moles of oxygen atoms = 4 moles of oxygen atoms.

B. Number of moles of O₃ molecules in 12g of O₃ = mass of O₃/molar mass of O₃

molar mass of O₃ = 48 g/mol

Number of moles of O₃ molecules = 12/48 = 0.25 moles of O₃

There are 3 atoms of oxygen in a molecule of O₃, therefore number of moles of oxygen atoms present in 12 g of O₃ = 3 * 0.25 = 0.75 moles of oxygen atoms

C. There are 5 oxygen atoms in a molecule of V₂O₅

0.1 mol of V₂O₅ contains 0.1 * 5 moles of oxygen atoms = 0.5 moles of oxygen atoms.

D. mass of 10 mL of H₂O = 10 g since the density of H₂O = 1 g/mL

molar mass of H₂O = 18.0 g/mol

Number of moles of H₂O in 10 g of H₂O = 10/18 = 0.56 moles

There are 1 atom of oxygen in a molecule of H₂O, therefore, number of moles of oxygen present in 10 g of H₂O = 0.56 * 1 = 0.56 moles of oxygen atoms

Therefore, 0.1 mole of V₂O₅ contains the least number of oxygen atoms.

6 0

3 years ago

Chem Muti Choice. Tell me the correct answer.

Alex787 [66]

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