Acetylene gas c2h2 undergoes combustion to form carbon dioxide and water when it is used in the oxyacetylene torch for welding. balance

The answer to your question is below Explanation: Reaction                             C₂H₂ (g) + O₂(g)   ⇒   CO₂ (g)   +  H₂O (g)                             Reactants         Elements         Reactants                                     2                       C                       1                                     2                       H                       2                                     2                       O                       3 This reaction is unbalance Reaction balanced                           2C₂H₂ (g) +   5O₂(g)   ⇒   4CO₂ (g)   +  2H₂O (g)                             Reactants         Elements         Reactants                                     4                       C                       4                                     4                       H                       4                                    10                       O                      10 Now, the reaction is balanced a) Calculate the molecular mass of acetylene and water Acetylene = (12 x 2) + (2) = 26 g Water = (1 x 2) + (1 x 16) = 18 g                            2(26) g of Acetylene  2(18) g of Water                                113 g  of Acetylene   x                                 x = (113 x (2 x 18)) / 2(26)                                 x = 4068 / 52                                x = 78. 2 g of water b)                2 moles of Acetylene  4 moles of carbon dioxide                    x moles of acetylene  1.10 moles of carbon dioxide                          x = (1.10 x 2) / 4                         x = 0.55 moles of acetylene

The answer to your question is below Explanation: Reaction                             C₂H₂ (g) + O₂(g)   ⇒   CO₂ (g)   +  H₂O (g)                             Reactants         Elements         Reactants                                     2                       C                       1                                     2                       H                       2                                     2                       O                       3 This reaction is unbalance Reaction balanced                           2C₂H₂ (g) +   5O₂(g)   ⇒   4CO₂ (g)   +  2H₂O (g)                             Reactants         Elements         Reactants                                     4                       C                       4                                     4                       H                       4                                    10                       O                      10 Now, the reaction is balanced a) Calculate the molecular mass of acetylene and water Acetylene = (12 x 2) + (2) = 26 g Water = (1 x 2) + (1 x 16) = 18 g                            2(26) g of Acetylene  2(18) g of Water                                113 g  of Acetylene   x                                 x = (113 x (2 x 18)) / 2(26)                                 x = 4068 / 52                                x = 78. 2 g of water b)                2 moles of Acetylene  4 moles of carbon dioxide                    x moles of acetylene  1.10 moles of carbon dioxide                          x = (1.10 x 2) / 4                         x = 0.55 moles of acetylene

a) 78.19 grams H2O b) 14.3 grams acetylene Explanation: Step 1: Data given Molar mass of acetylene = 26.04 g/mol Molar mass of H2O = 18.02 g/mol Step 2: The balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O Step 3: a. How many grams of water can form if 113g of acetylene is burned? Calculate moles of acetylene: Moles = mass / molar mass Moles = 113.0 grams / 26.04 g/mol Moles = 4.339 moles calculate moles of H2O For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O For 4.339 moles of acetylene we’ll have 4.339 moles H2O Calculate mass of H2O Mass H2O = 4.339 moles * 18.02 g/mol Mass H2O = 78.19 grams H2O b. How many grams of acetylene react if 1.10 mol of CO2 are produced? For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O For 1.10 mol CO2 we need 1.10/2 = 0.55 moles of acetylene Mass acetylene = 0.55 moles * 26.04 g/mol Mass acetylene = 14.3 grams acetylene

a) 78.19 grams H2O b) 14.3 grams acetylene Explanation: Step 1: Data given Molar mass of acetylene = 26.04 g/mol Molar mass of H2O = 18.02 g/mol Step 2: The balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O Step 3: a. How many grams of water can form if 113g of acetylene is burned? Calculate moles of acetylene: Moles = mass / molar mass Moles = 113.0 grams / 26.04 g/mol Moles = 4.339 moles calculate moles of H2O For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O For 4.339 moles of acetylene we’ll have 4.339 moles H2O Calculate mass of H2O Mass H2O = 4.339 moles * 18.02 g/mol Mass H2O = 78.19 grams H2O b. How many grams of acetylene react if 1.10 mol of CO2 are produced? For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O For 1.10 mol CO2 we need 1.10/2 = 0.55 moles of acetylene Mass acetylene = 0.55 moles * 26.04 g/mol Mass acetylene = 14.3 grams acetylene

Acetylene gas C2H2 undergoes combustion to form carbon dioxide and water when it is used in the oxyacetylene torch for welding. Balance the reaction and answer the following questions. C2H2(g)+O2(g) —> CO2(g)+H2O(g) a. How many grams of water can form if 113g of acetylene is burned? b. How many grams of acetylene react if 1.10 mol of CO2 are produced? PLEASE SHOW YOUR WORK! Explanation: Acetylene gas C2H2 undergoes combustion to form carbon dioxide and water when it is used in the oxyacetylene torch for welding. Balance the reaction and answer the following questions. C2H2(g)+O2(g) —> CO2(g)+H2O(g) a. How many grams of water can form if 113g of acetylene is burned? b. How many grams of acetylene react if 1.10 mol of CO2 are produced? PLEASE SHOW YOUR WORK!

Answer 6

C2H2 + 3O2 –> 2H2O + 2CO2 Using the molar ratios, we get that we get 8.68 mols of water for every 4.34 mols, so since the molar mass of water is 18 grams per mol, we multiply 18 * 8.68 = about 156.24 grams.

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