0.839
Our lower number of moles is .316 so we will use that. The total molar mass of PCl5 is 208.2. So we will multiply the lower number of moles by our total mass. .316mol X 208.2 g/mol = 65.8g <------ answer.
What mass of will be produced from the given masses of both reactants? 28.0 g/P4 / 123.9 g/P4 = 0.2260 moles/ P4 0.2260 moles /P4 * 4 moles PCl5/1 mole/P4=0.904 moles/PCl5 54.0 g/Cl2 / 70.9 g/Cl2 = .7616 moles/Cl2 .7616 moles / Cl2 *4 moles PCl5/10 moles Cl2=.30 This is about limiting reagents You need to use 2P + 5Cl2 --> 2PCl5 [or P4 + 10Cl2 --> 4PCl5 if you prefer] that tells you that 2P = 62g needs 5Cl2 = 355g Cl2 to react 69.3g
Assuming that you are combining the P4 with Cl2 and there is a suffiecient quantity of Cl2 for the P4 to completely react, you will first need a balanced equation which is P4 + 10Cl2 -> 4PCl5. From there, it's mostly stoichiometry. Take the 24g of P4, divide by the molar mass (123.88g/mol) to get the number of moles of P4 that you have (0.194). You then have to convert, using the balanced equation, from moles of P4 to moles of PCl5, in this case multiplying by 4. That will give you the number of moles of PCl5. The stoichiometry should look something like this 24.0 g P4 x (1 mol P4/123.88g P4) x (4 mol PCl5/1 mol P4).
The name of the compound PCl5 is phosphorus pentachloride.
The name for the molecular compound PCl5 is phosphorus pentachloride.
To determine how many moles of PCl5 can be produced from 58.0 g of Cl2, we first need to calculate the moles of Cl2. The molar mass of Cl2 is approximately 70.9 g/mol, so the number of moles of Cl2 is 58.0 g / 70.9 g/mol ≈ 0.819 moles. The balanced reaction for the formation of PCl5 from P4 and Cl2 is: P4 + 10 Cl2 → 4 PCl5. From this, we see that 10 moles of Cl2 produce 4 moles of PCl5, so 0.819 moles of Cl2 can produce (0.819 moles Cl2) * (4 moles PCl5 / 10 moles Cl2) ≈ 0.3276 moles of PCl5. Thus, approximately 0.328 moles of PCl5 can be produced.
Our lower number of moles is .316 so we will use that. The total molar mass of PCl5 is 208.2. So we will multiply the lower number of moles by our total mass. .316mol X 208.2 g/mol = 65.8g <------ answer.
55.0 g of Cl2 contains 55.0/35.45* or 1.551 gram atoms of chlorine. Each mole of PCl5 requires exactly 5 gram atoms of chlorine, as shown by the formula. Therefore, 1.551/5.000 or 0.310 moles of PCl5 can be formed, to the justified number of significant digits. *This number is the gram Atomic Mass of chlorine.
What mass of will be produced from the given masses of both reactants? 28.0 g/P4 / 123.9 g/P4 = 0.2260 moles/ P4 0.2260 moles /P4 * 4 moles PCl5/1 mole/P4=0.904 moles/PCl5 54.0 g/Cl2 / 70.9 g/Cl2 = .7616 moles/Cl2 .7616 moles / Cl2 *4 moles PCl5/10 moles Cl2=.30 This is about limiting reagents You need to use 2P + 5Cl2 --> 2PCl5 [or P4 + 10Cl2 --> 4PCl5 if you prefer] that tells you that 2P = 62g needs 5Cl2 = 355g Cl2 to react 69.3g
Assuming that you are combining the P4 with Cl2 and there is a suffiecient quantity of Cl2 for the P4 to completely react, you will first need a balanced equation which is P4 + 10Cl2 -> 4PCl5. From there, it's mostly stoichiometry. Take the 24g of P4, divide by the molar mass (123.88g/mol) to get the number of moles of P4 that you have (0.194). You then have to convert, using the balanced equation, from moles of P4 to moles of PCl5, in this case multiplying by 4. That will give you the number of moles of PCl5. The stoichiometry should look something like this 24.0 g P4 x (1 mol P4/123.88g P4) x (4 mol PCl5/1 mol P4).
Molar mass = 31+ 5(35.5) = 208.5 g/mole divide ... 35.6 g / 208.5 g/mole = 0.171 mole
The two compounds formed between PCl5 and ethanol are ethyl chloride (C2H5Cl) and diethyl ether (C4H10O). Ethyl chloride is produced when one mole of PCl5 reacts with one mole of ethanol, while diethyl ether is formed when two moles of ethanol react with one mole of PCl5.
If you mean elemental Phosphorous (As in, just a chunk of P), I believe that the reaction would go to Phosphorous pentachloride like this: P + 5 Cl--> PCl5 THe Lewis Dot structure works out that way, anyways.
Phosphorus pentachloride (PCL5)
PCl5 Phosphorous pentachlorideP for Phosphorous and 5 Cl for chloride= PCl5 College Chemistry student
The name of the compound PCl5 is phosphorus pentachloride.
The equation for phosphorus burning in the excess of chlorine is: 4 P (s) + 5 Cl2 (g) → 4 PCl5 (s)