It would be C
2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj
This is true, this isn't a question, it's a fact.
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Pressure caused by high temperatures are balanced by gravity
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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