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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!

Respuesta :

joseaaronlara

Answer:

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

Answer:

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

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