Since acetylene (C2H2) has a stoichiometry of 2 moles of acetylene to produce 2 moles of CO2, three moles of acetylene would produce 3 moles of CO2. The reaction with excess oxygen ensures that all the acetylene is fully converted to 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
CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq) Number of moles greater than 1 are prefixes of molecule formula's. Example: '2' in 2H2O(l)
To calculate the amount of C2H2 produced from H2O, we need to consider the stoichiometry of the reaction. The balanced equation for the reaction is 2H2O -> 2H2 + O2 -> 2C2H2. From 80 grams of H2O, we can calculate the amount of C2H2 produced using stoichiometry.
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.
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.
Since acetylene (C2H2) has a stoichiometry of 2 moles of acetylene to produce 2 moles of CO2, three moles of acetylene would produce 3 moles of CO2. The reaction with excess oxygen ensures that all the acetylene is fully converted to CO2.
The balanced chemical equation for the combustion of acetylene is: 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O From the equation, 2 moles of acetylene react with 5 moles of oxygen. Therefore, 12 moles of oxygen would react with 12/5 * 2 = 4.8 moles of acetylene.
2C2H2 + 5O2 ---->4CO2 + 2H2O so 1 mole of acetylene (ethyne) reacts with 2.5 moles of oxygen. Answer 2.5 moles
CaC2 + 2H2O ==> C2H2 + Ca(OH)2moles of C2H2 produced = moles of Ca(OH)2 produced (1:1 mole ratio is the stoichiometry)moles of Ca(OH)2 produced = 7 g x 1mol/74.1 g = 0.0945 molesmoles C2H2 = 0.0945 molesgrams C2H2 = 0.0945 moles x 26 g/mole = 2.46 grams = 2 grams to 1 sig fig
I have already balanced - 2C2H2+5O2->4CO2+2H2O what do I do next? 2 moles acetylene--------------5 moles oxygen O2 7 moles acetylene--------------x moles oxygen O2 x = 7 x 5/2 = 17,5
Two elements Carbon and Hydrogen in acetylene, C2H2, two atoms of each.
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
How many moles of CO2 are produced when 2.1 mol of C2H2 react?
CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq) Number of moles greater than 1 are prefixes of molecule formula's. Example: '2' in 2H2O(l)
The reaction between calcium carbide and water produces acetylene gas (C2H2) and calcium hydroxide (Ca(OH)2. When 1 mole of calcium carbide reacts with 2 moles of water, it produces 1 mole of acetylene and 1 mole of calcium hydroxide. The molar mass of calcium carbide is 64.1 g/mol, so the reaction would produce 26.04 grams of acetylene gas and 74.1 grams of calcium hydroxide from every 64.1 grams of calcium carbide reacted.
=26 x (pressure in kpa times 75.0)/(8.31 x 293)