Answer:
Mass of methane required in gram = 226.35 gram (Approx.)
Explanation:
Given:
Heat energy used = 12,567 kJ
Missing data;
Δcombustion = 891 kJ/mol
Find:
Mass of methane required in gram
Computation:
Number of mol methane burned = Heat energy used / Δcombustion
Number of mol methane burned = 12,571 / 891
Number of mol methane burned = 14.1088 mol
1 mol of methane in gram = 16.043 grams
So,
Mass of methane required in gram = Number of mol methane burned x 1 mol of methane in gram
Mass of methane required in gram = 14.1088 x 16.043
Mass of methane required in gram = 226.3474
Mass of methane required in gram = 226.35 gram (Approx.)
Answer: 0.0009 moles of octane were used up.
Explanation:
To calculate the moles :
Given mass of octane = 0.1 g
Molar mass of octane = 11.23 g/mol
Putting in the values we get:

Thus 0.0009 moles of octane were used up.
Answer:
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Explanation:
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Answer:
0.095 moles of O₂ are left over.
Explanation:
First of all, state the balanced reaction:
2NO + O₂ → 2NO₂
We determine moles of each reactant:
20.2 g . 1mol / 30g = 0.673 moles of NO
13.8g . 1mol / 32g = 0.431 moles of oxygen
Oxygen is the excess reactant. Let's see.
For 2 moles of NO I need 1 mol of O₂
Then, for 0.673 moles of NO I may use (0.673 .1) /2 = 0.336 moles
I have 0.431 moles of O₂ and I only need 0.336 mol. According to reaction, stoichiometry is 2:1.
In conclussion, the moles of excess reactant that will be left over:
0.431 - 0.336 = 0.095 moles