For this reaction, the stoichiometry indicates that 4 moles of CO2 are produced for every 2 moles of C2H2 reacted. The molar mass of CO2 is approximately 44 g/mol. Therefore, you can calculate the mass of CO2 produced by converting the moles of CO2 into grams.
First you need to work out the balanced equation. 2C2H2 + 5O2 -----> 4CO2 + 2H2O From this we see that 2 moles of acetylene produces 4 moles of carbon dioxide. 1 mole of carbon dioxide is 12 + 16 + 16 = 44g (adding the mass numbers of the component elements). If 1 mole = 44g then 4 mole = 176g
In the reaction 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l), the entropy decreases. This is because the reactants consist of gaseous molecules, which have higher entropy due to their greater freedom of movement, while the products include liquid water, which has lower entropy. Additionally, there is a reduction in the number of gas molecules from 7 (2 C2H2 + 5 O2) to 4 (4 CO2), further contributing to the decrease in disorder. Overall, the transition from gas to liquid and the reduction in the number of gas molecules results in a net decrease in entropy.
FeedbackThe correct answer is: 2C2H2 + 5O2 4CO2 + 2H2O.
During oxyacetylene gas welding, acetylene gas and oxygen are mixed in a torch and ignited to produce a high-temperature flame. The heat from this flame melts the metal being welded, while the oxygen assists in the combustion process by providing additional heat and reacting with the metal to form an oxide layer that is then removed as slag.
To determine the number of grams of water formed, we need to calculate the moles of butanol (C4H9OH) and then use the balanced chemical equation to find the moles of water produced in the combustion reaction. From there, we convert moles of water to grams. The balanced equation for the combustion of butanol is C4H9OH + 6O2 → 4CO2 + 5H2O.
The balanced equation for the complete combustion of ethyne (C2H2) is: 2C2H2 + 5O2 -> 4CO2 + 2H2O
It is the balanced equation for the combustion of acetylene (or ethyne).
The balanced equation for the complete oxidation of acetylene (C2H2) burning in air is: 2C2H2 + 5O2 -> 4CO2 + 2H2O. This equation shows that two molecules of acetylene react with five molecules of oxygen to produce four molecules of carbon dioxide and two molecules of water.
The balanced chemical equation for the combustion of acetylene is: 2C2H2 + 5O2 → 4CO2 + 2H2O. This means 2 moles of C2H2 produce 4 moles of CO2. Therefore, 1.3 moles of C2H2 will produce 2.6 moles of CO2, which is equivalent to approximately 84.8 grams of CO2.
First you need to work out the balanced equation. 2C2H2 + 5O2 -----> 4CO2 + 2H2O From this we see that 2 moles of acetylene produces 4 moles of carbon dioxide. 1 mole of carbon dioxide is 12 + 16 + 16 = 44g (adding the mass numbers of the component elements). If 1 mole = 44g then 4 mole = 176g
The balanced equation for the combustion of acetylene (C2H2) in air to form carbon dioxide (CO2) and water (H2O) is: 2C2H2 + 5O2 -> 4CO2 + 2H2O
The reaction is: 2 C2H2 + 5 O2 -------------- 4 CO2 + 2 H2O.
The C2H2 us being oxidized, electrons stripped away, and the oxygen is being reduced, accepting electrons and in this case, carbon and hydrofen are coming along for the ride.
there are many possible combinations, as incomplete combustion refers only to the combustion of a fuel whereby not all of a fuel's carbon and hydrogen is converted to carbon dioxide and water, usually other products involve carbon particulates (C) or carbon monoxide (CO) as such two possible equations are:C2H4 + O2 => C2 + 2H20C2H4 + 2O2 => 2CO + 2H2O
In the reaction 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l), the entropy decreases. This is because the reactants consist of gaseous molecules, which have higher entropy due to their greater freedom of movement, while the products include liquid water, which has lower entropy. Additionally, there is a reduction in the number of gas molecules from 7 (2 C2H2 + 5 O2) to 4 (4 CO2), further contributing to the decrease in disorder. Overall, the transition from gas to liquid and the reduction in the number of gas molecules results in a net decrease in entropy.
2C2H2(g) + 5O2(g) = 2H2O(g) + 4CO2(g) or, 2 Acetylene molecules + 5 Oxygen molecules = 2 water molecules + 4 Carbon dioxide molecules (+ energy)
ΔS is positive and G is negative at all temp.Which of the following is true for the gas phase reaction shown below? 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g), ΔH = -2511 kJΔS is negative and ΔG is negative at low temperatures.