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The amount, in mg, of CO present in the room will be 191,520 mg.
<h3>Stoichiometric problem</h3>The concentration of the gas in the room is 5.7 x mg/cm3.
The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.
We can obtain the volume of the room as:
350 x 300 x 320 = 33,600,000 cm3
The concentration is in mg/cm3, meaning that it is mass/volume.
Thus:
mass = concentration x volume = 5.7 x mg/cm3 x 33,600,000 cm3
= 191,520 mg
The mass of CO in the room is 191,520 mg
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer: 714 g Al2O3
Explanation: Solution attached
First convert mass of O2 to moles
Do the mole ratio between O2 and Al2O3 from the balanced equation.
Convert moles of Al2O3 to mass using its molar mass.
