aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is mixed with 13.9 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
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Ответ:
There is 9.6 grams of CO2 produced
Explanation:
Step 1: Data given
Mass of methane = 5.50 grams
Molar mass of methane = 16.04 g/mol
Mass of oxygen = 13.9 grams
Molar mass of oxygen = 32.0 g/mol
Step 2: The reaction
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Step 3: Calculate number of moles
Moles = mass / molar mass
Moles methane = 5.50 grams / 16.04 g/mol
Moles methane = 0.343 moles
Moles oxygen = 13.9 grams / 32.0 g/mol
Moles oxygen = 0.434 moles
For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O
O2 is the limiting reactant. It will completely react (0.434 moles).
There will react 0.434/2 = 0.217 moles CH4
There will remain 0.343-0.217 = 0.126 moles CH4
There will be produced 0.434 moles of H2O and
0.434/2 =0.217 moles of CO2
Step 4: Calculate mass of products
Mass = moles * molar mass
Mass CO2 = 0.217 moles ¨44.01 g/mol
Mass CO2 = 9.6 grams
Mass H2O = 0.434 moles * 18.02
Mass H2O = 7.8 grams
Ответ:
A
Explanation:
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