Can you please help me with the following problems oftheoretical yield (34). I a
ID: 681831 • Letter: C
Question
Can you please help me with the following problems oftheoretical yield (34). I am desperate and can not get anyone tohelp me please!!! 1) How many moles of HCl could be formed from 5.31 moles ofPCl3 and 10.18 moles of water in the followingreaction? PCl3 + 3H2O ---> 3HCl + H3PO3 2) How many moles of Ba3(PO4)2could be formed from 4.31 moles of Ba(NO3)2and 3.58 moles of Na3PO4 in the followingreaction? 3 Ba(NO3)2 + 2 Na3PO4 ----> Ba3(PO4)2 + 6NaNO3 3) How many grams of Ca3N2 can be formedfrom 4.68 g of calcium and 0.36 g of nitrogen in the followingreaction? 3 Ca + N2 ---> Ca3N2 4) How many grams of SO3 can be formed from 12.07 g ofSO2 and 2.46 g of O2 in the followingreaction? 2 SO2 + O2 -----> 2 SO3Can you please help me with the following problems oftheoretical yield (34). I am desperate and can not get anyone tohelp me please!!! 1) How many moles of HCl could be formed from 5.31 moles ofPCl3 and 10.18 moles of water in the followingreaction? PCl3 + 3H2O ---> 3HCl + H3PO3 2) How many moles of Ba3(PO4)2could be formed from 4.31 moles of Ba(NO3)2and 3.58 moles of Na3PO4 in the followingreaction? 3 Ba(NO3)2 + 2 Na3PO4 ----> Ba3(PO4)2 + 6NaNO3 3) How many grams of Ca3N2 can be formedfrom 4.68 g of calcium and 0.36 g of nitrogen in the followingreaction? 3 Ca + N2 ---> Ca3N2 4) How many grams of SO3 can be formed from 12.07 g ofSO2 and 2.46 g of O2 in the followingreaction? 2 SO2 + O2 -----> 2 SO3
Explanation / Answer
Hello, The key to solving these problems is to identify the mol ratiobetween your reactants and products of interest. Then, you willcalculate the amount of product you can make with each of thereactants separately while assuming the other reagents are infinitein supply. Then you look to see which of your reactants producesless product and it is the limiting reagent. Think of it like this: You have x amount of peanut butter and yamount of jelly. If you were told to see how many sandwiches youcould make, then you are limited by the whichever of youringredients runs out first. Apply the same principle to thefollowing problems. 1). a) Assume you have an infinite amount of H2O. Looking atthe balanced equation, you can see that for each mol of PCl3, youhave 3 moles of HCl. Then to get the number of moles of HCl fromyour number of moles of PCl3, just do the following conversion: (5.31 mol of PCl3)*(3 mol HCl/1mol PCl3) = 15.93mol HCl b) Assume you have an infinite amount ofPCl3.Apply the same concept when solving for water. Thebalanced equation shows that for every 3 moles of water, you have 3moles of HCl: (10.18mol of H2O)*(3mol HCl/3mol H2O) = 10.18mol HCl ---Your limiting reagent is PCl3 and you can only make 10.18molHCl 2) Apply the same methods as in step 1) a) Assume you have an infinite amount ofNa3PO4(4.31mol Ba(NO3)2)*[(1mol Ba3(PO4)2)/3 molBa(NO3)2] = 1.437mol Ba3(PO4)2 b) Assume you have an infinite amount ofBA(NO3)2(3.58mol Na3PO4)*[(1mol Ba3(PO4)2)/2molNa3Po4] = 1.79mol Ba3(PO4)2 ---Your limiting reagent is Ba(No3)2 and you can only make 1.79molBa3(PO4)2 3) Solving these problems where you are given massvalues instead of mole values, you can just convert your masses torespective mole values and then doing the same as in part 1) and 2)such that: a) Assume you have an infinite amount of N2 You have 4.68g of Ca, so convert that to moles: (4.68g of Ca)*(1mol Ca/40.1g) = 0.117mol Ca. Now that you have moles of Ca, use the balanced equation to findthe mol ratio between Ca and Ca3N2 which is 3Ca to 1 Ca3N2: (0.117mol Ca)*(1mol Ca3N2/3mol Ca) = 0.039mol Ca3N2 Convert molesof Ca3N2 to mass: (0.039mol Ca3N2)*(64g Ca3N2/1mol Ca3N2) = 2.496g Ca3N2 b) Assume you have an infinite amount ofCaRepeat the same step as in 3a): (0.36g N2)*(1mol N2/28g) = 0.0129mol N2 The balanced equationshows that the mol ratio between N2 and Ca3N2 is 1:1 so you get thefollowing: (0.0129mol N2)*(1mol Ca3N2/1mol N2) = 0.0129mol Ca3N2 Convert molesof Ca3N2 to mass: (0.0129mol Ca3N2)*(64g/1mol Ca3N2) = 0.8256g of Ca3N2 ---Your limiting reagent is N2 and you can only make 0.8256g ofCa3N2 4) Repeat the same thing you did in step 3) a) Assume you have an infinite amount ofO2 (12.07g SO2)*(1mol SO2/64.1g SO2) = 0.188mol SO2 Your mol ratio says you get 2 SO3 for every 2 SO2: (0.188mol SO2)*(2mol SO3/2 SO2) = 0.188 mol SO3 Convert your moles to mass (0.188mol SO3)*(80.1g SO3/1mol SO3)=15.06g SO3 b) Assume you have an infinite amount of SO2 Repeat the steps in 4a (2.46g O2)*(1mol O2/32g O2)*(2mol SO3/1mol O2)*(80.1g SO3/1mol SO3)= 12.32g SO3 ---Your limiting reagent is O2 and you can only make 12.32g SO3