0.449 mol
The relative number of moles of hydrogen to moles of oxygen that react to form water represents the stoichiometry of the chemical reaction according to the balanced equation. This relationship reflects the proportions in which the reactants combine to form the products.
Carbon dioxide is the limiting reagent.
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 number of moles in exactly 64 grams of oxygen (O2) is two.
The number of moles is 0,4375.
0.678 - 0.682
After the reaction is complete, all of the calcium will react with 3.165 mol (since calcium and oxygen react in a 1:1 ratio) of the oxygen gas. This means that there will be 4.00 mol - 3.165 mol = 0.835 mol of oxygen gas left over.
Since oxygen is diatomic it requires 2 moles of oxygen.
The relative number of moles of hydrogen to moles of oxygen that react to form water represents the stoichiometry of the chemical reaction according to the balanced equation. This relationship reflects the proportions in which the reactants combine to form the products.
For every mole of oxygen consumed in the reaction 2H2 + O2 -> 2H2O, two moles of water are produced. Therefore, if 0.633 moles of oxygen are consumed, the number of moles of water produced would be 2 x 0.633 = 1.266 moles.
Carbon dioxide is the limiting reagent.
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.
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
Based on the balanced equation: 2Ca + O2 --> 2CaO Two moles of calcium will be consumed for every one mole of oxygen used. This means that if 6.33 mol Ca is used, half this, or 3.165 mol O2 will be needed (use a simple proportion or dimensional analysis). Subtract the amount of O2 used from 4.00.
For every mole of potassium chlorate that decomposes, three moles of oxygen are produced. Therefore, if 7.5 moles of potassium chlorate decompose, 22.5 moles of oxygen would be produced (7.5 moles x 3).
The number of moles in exactly 64 grams of oxygen (O2) is two.
The number of moles is 0,4375.