P4(s)+10Cl2(g)→4PCl5(g) 0.646 mol of PCl5 can be produced from 20g of P4 , 0.3155 mol of...

Question

Question

P₄(s)+10Cl₂(g)→4PCl₅(g)

0.646 mol of PCl₅ can be produced from 20g of P₄ , 0.3155 mol of PCl₅ can be produced from 56 g of Cl₂

What mass of PCl₅ will be produced from the given masses of both reactants?

Answer 1

Answer #1

06·06·2006 lo c veu Mal.wayな 123.83s amale mans P o.GVSCX 203.2y ン 0.1887 rnoe 0.3154-mole-t-4- 65.47

Answer 2

Similar Homework Help Questions

What mass of PCl5 will be produced from the given masses of both reactants? P4+10Cl2 —>...

What mass of PCl5 will be produced from the given masses of both reactants? P4+10Cl2 —> 4PCl5 20 g of P4 and 56 g of Cl2
P4(s)+10Cl2(g)→4PCl5(g) What mass of PCl5 will be produced from the given masses of both reactants? Express...

P4(s)+10Cl2(g)→4PCl5(g) What mass of PCl5 will be produced from the given masses of both reactants? Express your answer to three significant figures and include the appropriate units. 0.872 mole and 0.310 reactants 0.872 moleWhat is the percent yield if the actual yield from this reaction is 347 g ? Express your answer to three significant figures and include the appropriate units.
P4(s)+10Cl2(g)→4PCl5(g) Now consider a situation in which 29.0 g of P4 is added to 55.0 g...

P4(s)+10Cl2(g)→4PCl5(g) Now consider a situation in which 29.0 g of P4 is added to 55.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations: 1. Calculate the number of moles of PCl5 that can be produced from 29.0 g of P4 (and excess Cl2). 2. Calculate the number of moles of PCl5 that can be produced from 55.0 g of Cl2 (and excess P4). 3. Then, compare the...

How many moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2)...

How many moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2) P4(s)+10Cl2(g)→4PCl5(g)
1. How many moles of PCl5 can be produced from 30.0 g of P4 (and excess...

1. How many moles of PCl5 can be produced from 30.0 g of P4 (and excess Cl2)? 2.How many moles of PCl5 can be produced from 59.0 g of Cl2 (and excess P4)? 3.What mass of PCl5 will be produced from the given masses of both reactants?
3)Use the balanced equation: P4 + 10Cl2 → 4PCl5. When 23.0 g P4 and 16.0 g...

3)Use the balanced equation: P4 + 10Cl2 → 4PCl5. When 23.0 g P4 and 16.0 g Cl2 react, how many grams of PCl5 are produced? Which reactant is limiting and which is in excess?

Determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=? by making use of the following information:...

Determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=? by making use of the following information: P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019 PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2
What mass of PCl5 will be produced from the given masses of both reactants? Calculate the...

What mass of PCl5 will be produced from the given masses of both reactants? Calculate the number of moles of PCl5 that can be produced from 23.0 g of P4 (and excess Cl2). 0.743 mole Calculate the number of moles of PCl5 that can be produced from 53.0 g of Cl2 (and excess P4). 0.299 mole
What mass of PCl5 will be produced from the given masses of both reactants? 28.0 g...

What mass of PCl5 will be produced from the given masses of both reactants? 28.0 g of P4 and 59.0 g of Cl2

In the reaction below, what volume of PCl5(g) will be produced by the reaction of 18.9...

In the reaction below, what volume of PCl5(g) will be produced by the reaction of 18.9 L of chlorine gas with excess phosphorus (P4) if all gas volumes are measured at STP? P4(g) + Cl2(g) => PCl5(g)(unbalanced) Enter your answer with two decimal places and no units. The answer should be 7.56, but I keep getting 6.92. Balanced P4 + 10Cl2 ----> 4PCl5 1 atm * 18.9L = n (.0821)*(298) 0.7725*(4/10)*22.4 =6.92 , I multiplied it times 22.4 because its...