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?
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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