This enthalpy of reaction is 560 joules, for liquid water at 25 oC.
There are 18 moles of water produced in the reaction. This is determined by the stoichiometry of the balanced chemical equation, which shows that for every 2 moles of C8H18 consumed, 18 moles of H2O are produced.
To determine how many moles of CH4 (methane) are produced along with 11 moles of water, we need the balanced chemical equation for the reaction. In the case of methane production from a reaction like the one involving carbon dioxide and hydrogen (CO2 + 4H2 → CH4 + 2H2O), every mole of CH4 produced yields 2 moles of water. Therefore, if you have 11 moles of water, you would produce 5.5 moles of CH4.
No, this would make 5 moles. This is because water is H2O. This means that for each oxygen molecule used, there will be 2 hydrogen molecules used. In the given equation Only 5 moles of oxygen could be used to pair with all 10 moles of hydrogen, therefore giving you an excess of 5 oxygen molecules.
To determine the moles of H2O required for the reaction with 0.24 moles of Cl2, we first need the balanced chemical equation. For example, in the reaction of chlorine gas with water, Cl2 + H2O → HCl + HOCl, one mole of Cl2 reacts with one mole of H2O. Therefore, 0.24 moles of Cl2 would require 0.24 moles of H2O.
Two moles of water are produced.
Assuming water vapor is the limiting reactant and undergoes a complete reaction, the stoichiometry of the reaction is as follows: 2 moles of water vapor yield 2 moles of hydrogen gas. Therefore, 174.82 moles of hydrogen would require 174.82 moles of water vapor to react.
There are 18 moles of water produced in the reaction. This is determined by the stoichiometry of the balanced chemical equation, which shows that for every 2 moles of C8H18 consumed, 18 moles of H2O are produced.
To determine the limiting reactant, you must compare the moles of each reactant to the stoichiometry of the reaction. The balanced equation is essential to determine the ratio of moles needed for the reaction. In this case, 3.00 moles of calcium and 8.00 moles of water are given, and you can find which reactant limits the reaction by finding out which reactant would require more moles for complete reaction based on the stoichiometry.
To determine the moles of water produced from the reaction of 6.00 grams of propane, first calculate the moles of propane using its molar mass. Then, use the balanced chemical equation to find the moles of water produced based on the stoichiometry of the reaction.
The balanced chemical equation for the reaction between sodium and water is: 2 Na + 2 H2O -> 2 NaOH + H2 Since 2 moles of water are needed to react with 2 moles of sodium, 1 mole of water is needed to react with 1 mole of sodium. Therefore, 2.5 moles of sodium will require 2.5 moles of water for the reaction.
No, this would make 5 moles. This is because water is H2O. This means that for each oxygen molecule used, there will be 2 hydrogen molecules used. In the given equation Only 5 moles of oxygen could be used to pair with all 10 moles of hydrogen, therefore giving you an excess of 5 oxygen molecules.
The balanced equation for the synthesis of water from hydrogen and oxygen is: 2H2 + O2 --> 2H2O. If you have 1.4 moles of hydrogen, you will produce 1.4 moles of water because the molar ratio of hydrogen to water in this reaction is 2:2, which is the same as 1:1, which is the same as 1.4:1.4.
If the reaction is not specified, we can't determine the exact moles of NO formed from NO2 based on this information alone. The reaction and stoichiometry are needed to calculate the moles of NO produced from 8.44 moles of NO2.
The balanced chemical equation for the reaction between oxygen and hydrogen is2H2 + 02 -> 2H2OThus 2.2 moles of oxygen reacts with 4.4 moles of hydrogen to form 4.4 moles of steam (water in gaseous state).The mass of H2O obtained is thus 4.4 x 18.0 = 79.2g.
The balanced chemical equation for the reaction between ammonia (NH3) and water (H2O) is: 4NH3 + 5O2 → 4NO + 6H2O. This means that for every 4 moles of ammonia, 6 moles of water are produced. Therefore, if 2 moles of ammonia are used, 3 moles of water vapor can be produced.
During boiling the chemical composition is not changed.
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.