Phosphorous pentoxide, p2o5(s), is produced from the reaction between pure oxygen and pure phosphorous (p, solid). what is the volume of oxygen (in m3) that is used to complete a reaction that yields 6.92 kilograms of p2o5(s), when carried out at 396.90°c and 606.1 mmhg? (assume ideal-gas behaviour)

4190.22 L = 4.19 m³.

Explanation:

For the balanced reaction:

2P₂ + 5O₂ ⇄ 2P₂O₅.

It is clear that 2 mol of P₂ react with 5 mol of O₂ to produce 2 mol of P₂O₅.

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:

Using cross multiplication:

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: PV = nRT.

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.

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !

4- c, because all matter stays the same no matter what. it is still there. 5- b, because matter is known as mass. 6-c because, flammability is a chemical property. malleability is changing the shape of an object.