The metric prefix name for 1/100 is centimeters.
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In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,
The chemical equation is:
CH₄ + 2O₂ → CO₂ + 2H₂O
First, we calculate the moles of methane present using:
Moles = mass / molecular mass
Moles = 20 / 16
Moles = 1.25
Next, we may observe from the chemical equation that the molar ratio between methane and oxygen is 1 : 2
So the moles of oxygen required are 2 x 1.25
2.5 moles of oxygen required
Mass = moles * molecular mass
Mass = 2.5 * 32
Moles = 80
C. 80 grams O₂
