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?
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3)Use the balanced equation: P4 + 10Cl2 → 4PCl5.
When 23.0 g P4 and 16.0 g...
P4(s)+10Cl2(g)→4PCl5(g) Now consider a situation in which 29.0 g of P4 is added to 55.0 g...
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...
P4(s)+10Cl2(g)→4PCl5(g) 0.646 mol of PCl5 can be produced from 20g of P4 , 0.3155 mol of...
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?
Calculations involving a limiting reactant Now consider a situation in which 23.0 g of P4 is...
Calculations involving a limiting reactant Now consider a situation in which 23.0 g of P4 is added to 50.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations: Calculate the number of moles of PCl5 that can be produced from 23.0 g of P4 (and excess Cl2). Calculate the number of moles of PCl5 that can be produced from 50.0 g of Cl2 (and excess P4). Then, compare...
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
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)
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.
Calculate the work done when 5.81 g of Cl2(g) reacts with excess P4(s) at 22.0°C, according...
Calculate the work done when 5.81 g of Cl2(g) reacts with excess P4(s) at 22.0°C, according the the following equation: P4(s) + 10Cl2(g) → 4PCl5(s) Express your answer to three significant figures.
6)Using the balanced equation: SiO2 + 6HF → H2SiF6 + 2H2O, if 40.0 g SiO2 and...
6)Using the balanced equation: SiO2 + 6HF → H2SiF6 + 2H2O, if 40.0 g SiO2 and 40.0 g HF react, determine: a) how many grams of H2SiF6 are produced, b) the limiting reactant, c) the excess reactant, d) the left over mass of your excess reactant e) the percent yield if 45.8 g H2SiF6 was produced in lab.
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...
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