How many molecules of PBr_5 are in 65.0g PBr_5? H_2SO_4 + K rightarrow K_2SO_4 +

Question

How many molecules of PBr_5 are in 65.0g PBr_5? H_2SO_4 + K rightarrow K_2SO_4 + H_2 How many grams of K will complete the reaction with 56.6 g H_2SO_4? If we start with 39.54g K and react it with an excess of H_2SO_4, what would be the volume of the hydrogen produced at STP? Using the data in How many molecules of the potassium sulfate will be produced? MoCl_5 (aq) + AgNO_3 (aq) rightarrow AgCl + Mo(NO_3)_5 If we begin with 46.43g MoCl_5, how many grams of AgNO_3 will be needed to complete the reaction? Using the data from what would be the mass of the precipitate produced from this reaction? Using the data from again tell how many formula units of the product that remains aqueous would be produced in this reaction. C_4H_10O_2 + O_2 rightarrow CO_2 + H_2O If we completely react 74.6g C_4H_10O_2, how many liters of O_2 gas(at STP would complete the reaction? If we react 79.4L of O_2 gas(at STP) with excess C_4H_10O_2, how many molecules of H_2O would be produced? If we react 2.487 times 10^21 molecules of C_4H_10O_2 with excess O_2, what would be the volume of CO_2 produced at STP?

Explanation / Answer

To calculate the number of molecules of a given compound, initially we need to calculate the number of moles and multiply the obtained number of moles with the avogadro number.

For calculating this number of moles: Given weight of PBr5 = 65 g

Molecular Weight of PBr5 = (Atomic Weight of P) + (5* Atomic weight of Br) = 31+(5*80) =431 g/gmol

Number of moles = weight of PBr5/Molecular Weight of PBr5 = 65/431 = 0.151 moles

Number of molecules = 0.151 * 6.023*1023 = 0.909 *1023 molecules of PBr5 are in 65 g of PBr5

                                 

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.