Answer:
12 moles of H₂O are formed in this combustion.
Explanation:
First of all, think the reaction:
2CH₃OH (l) + 3O₂ (g) → 2CO₂ (g) + 4H₂O (g)
Ratio in the reactants is 2:3, so 2 mol of methanol need 3 mol of oxygen to react. Then 8 mol of CH₃OH, will need (8.3)/2 = 12 moles of O₂
We have 9 moles of O₂, so this is the limiting reactant.
3 mol of oxygen produce 4 mol of water
Then, 9 mol of oxygen will produce ( 9 .4)/3 = 12 moles
Answer: 287.8 cm3
Explanation:
Given that:
Initial volume of gas V1 = 350 cm3
Initial pressure of gas P1 = 740 mmHg
New volume V2 = ?
New pressure P2 = 900 mmHg
Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
740 mmHg x 350 cm3 = 900mmHg x V2
V2 = (740 mmHg x 350 cm3) /900mmHg
V2 = 259000 mmHg cm3 / 900mmHg
V2 = 287.8 cm3
Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.
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