a) Balance the equation. b) How many moles of H3PO4 would be produced from 108 grams of water? Answer= 4 mol H3PO4 P2O5 + 3 H2O 2 H3PO4 108 g x mol

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1 20- Given the equation P2O5 + H2O H3PO4 a) Balance the equation. Phosphorus Pentoxide + Water Phosphoric Acid Reaction Type: Synthesis b) How many moles of H3PO4 would be produced from 108 grams of water? To solve this Mole-Mass stoichiometry problem we have firstly balance the unbalanced equation. Herein in this problem we identify that we must find (x) number of moles of H3PO4 that can be produced from 108 grams of water. Secondly, we write the identified elements below the relevant formulas in the balanced equation as follow: 108 g x mol Thirdly, we find the GFM of H 2O [ 2H+1O= (2x1) + (1x16)=18 g/mol ] to use it in the factor-label method (dimensional analysis) based on our knowledge of chemistry as follow: x mol H3PO4 = (108 g H 2O) x (1 mol H2O (18 g H20) x (2 mol H 3PO 4) ( 3 mol H2O) = 4 mol H3PO4 from GFM from the balanced equation Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do the mathematical calculations to get the final answer. Answer= 4 mol H3PO4

2 c) How many grams of H3PO4 would be produced from moles of P2O5? To solve this Mass-Mole stoichiometry problem we have firstly balance the unbalanced equation. Herein, in this problem we identify that we must find (x) number of grams of H3PO4 that can be produced from moles of P2O5. Secondly, we write the identified elements below the relevant formulas in the balanced equation as follow: mol x grams Thirdly, we find the GFM of H3PO4 [ (3H+1P+4 O)= (3x1) + (1x31)+(4x16)=98 g/mol ] to use it in the factor-label method (dimensional analysis) based on our knowledge of chemistry as follow: x g H3PO4 = (0.400 mol P 2O 5) x (2 mol H3PO4 (1 mol P 2O 5) x (98 g H 3PO 4) ( 1 mol H3PO4) = 78.4 g H3PO4 from the balanced equation from GFM Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do the mathematical calculations to get the final answer. Answer= 78.4 g H3PO4 d) How many grams of water would be needed to react with 71.0 grams of P2O5? To solve this Mass-Mass Stoichiometry problem we must firstly balance if the equation is unbalance; identify what we are given and what we are asked to answer. Herein we identify that we must find (x) number of grams of H2O that will be needed to react with grams of P2O5. Secondly, we write the identified elements below the relevant formulas in the balanced equation as follow: 71.0 grams x grams Thirdly, we find the GFM of P2O5 and H2O as follow: 2 P=2x31=62 2 H= 2x1=2 5 O= 5x16=80 1 O= 1x16=16 142g/mol 18g/mol

3 Fourthly, we use factor-label method (dimensional analysis) based on our knowledge of chemistry as follow x g H2O = 71.0 g P2O5 x (1 mol P 2O 5) x (3 mol H 2O) x (18 g H 2O) = 27 g H2O (142 g P 2O 5) (1 mol P 2O 5) (1 mol H 2O) from GFM from the balanced equation from GFM Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do the mathematical calculations to get the final answer. Answer = 27 g H2O 21- If 100 grams of water reacts with 100 grams of calcium carbide according to the following reaction: CaC 2 (s) + 2 H 2O (l) Ca(OH) 2 (aq)+ C 2H 2 (g) a) Which reactant is the limiting reactant (reagent)? To solve this problem, firstly, we will find the GFM of both reactants and then convert the given grams of both reactants into moles. GFM of CaC 2 is [ 1Ca +2C= (1x40)+(2x12)= 64g/mol] GFM of H 2O is [ 2H+1O= (2x1) + (1x16)=18 g/mol ] mole of CaC 2 = 100 g of CaC 2 x 1mole of CaC 2 = 1.56 moles CaC 2 64 g of CaC 2 and moles of H 2O = 100 g H 2O x 1mole H 2O = 5.56 moles H 2O 18 g H 2O Secondly, we will divide calculated moles of the both reactants by the smallest number (1.56) 1.56 moles CaC 2 = 1 moles CaC 2 and 5.56 moles H 2O = 3.56 moles H 2O Thirdly we will write the calculated moles ratio right below the respective reactants in the given equation. 1 CaC 2 (s) + 2 H 2O (l) Ca(OH) 2 (aq)+ C 2H 2 (g) 1 mole 3.56 mole Per the given balanced chemical equation 1 mole of CaC 2 (s) reacts with 2 moles of H 2O (l). However, the calculated ratio of moles clearly shows that for the 1 mole of CaC 2 there are 3.56 moles of water available, which are way more than what the reaction needs. Therefore, it is obvious that the CaC 2 is the limiting reactant because after its consumption the chemical reaction will stop.

4 b) If 100 grams of calcium carbide yields 28.3 g of C 2H 2 g, what is the percent yield? CaC 2 (s) + 2 H 2O (l) Ca(OH) 2 (aq)+ C 2H 2 (g) To solve this problem, firstly we will find the GFM of CaC 2 [40+(2x12) = 64 g/mole and C 2H 2=[(2x12) +2x1)=26 g/mole] According to the equation 64 grams of CaC 2 produces 26 g of C 2H 2. The given100g of CaC 2 should produce (theoretical yield); 26 g C2H2 x g of C 2H 2 (theoretical yield) = 100 g CaC2 = g C2H2. However, the actual yield is 64 g CaC2 only g C 2H 2. Percent error = % error = [Measured value accepted value] [ accepted value ] 100 [28.3 g g] [ g ] = 64. Therefore, the % yield is only 64% 100 = 30.34% To double check whether our answer is correct or not we can do the following: Hence, the answer 69.66% yield is correct g 100 = 69.66% g 22- According to the following reaction 3NO2 (g)+ H2O (l)= 2HNO3 (l)+ NO (g) If you are given 28 g of NO 2 g and 18 g of water, which reactant is acting as the limiting reactant? To solve this problem, firstly, we will find the GFM of both reactants and then convert the given grams of both reactants into moles. GFM of NO 2 is [ 1N +2 O= (1x14)+(2x16)= 46g/mol] GFM of H 2O is [ 2H+1 O= (2x1) + (1x16)=18 g/mol ] mole of NO 2 = 28 g of NO2 x 1mole of NO2 = 0.61 moles NO2 46 g of NO2 and moles of H 2O = 18 g H 2O x 1mole H 2O = 1 moles H 2O 18 g H 2O

5 Secondly, we will write the calculated moles ratio right below the respective reactants in the given equation. 3NO2 (g) + H2O (l)= 2HNO3 (l)+ NO (g) 0.61 mol 1 mol Per the given balanced chemical equation 1 mole of H2O reacts with 3 moles of NO2 (g). However, the calculated ratio of moles clearly shows that for 1 mole of H2O there are 0.61 moles of NO2 (g) available, which are way less than what the reaction needs. Therefore, it is obvious that the NO2 (g) is the limiting reactant because after its consumption the chemical reaction will stop.