Given the balanced equation
Kr + 3F2 --> KrF6
In order to find how many moles of F2 are needed to produce 3.0 moles of KrF6, we must convert from moles to moles (mol --> mol conversion).
3.0 mol KrF6 * 3 molecules F2 = 9.0 mol F2
--------- 1 molecule F2
the required equation is HgCl2+4KI>>2KCl+K2HgI4. according to stoichiometric calculations 4 moles of KI gives 1 mole of k2HgI4 THEREFORE 0.4 moles of K2HgI4 requires----- ? 0.4 moles x 4 moles/1 mole=1.6 moles therefore 1.6 moles of KI is required to produce 0.4 moles of K2HgI4
800 g oxygen are needed.
They are called coefficients. They represent the number of moles of that element required for the reaction to be completed as written in relation to the number of moles for the other elements. They can also be considered the number of molecules required (different from moles), but most combustion reactions include 1/2 as a coefficient for O2. So don't get confused.
yes the number of moles is the the number before the substance for example if the equation is balanced and you want to find H2O and it appears like 3H2O in the equation then their are 3 moles of H2O
You can make a simple balance. There are (12.36 * 3) moles of H You have 2*H to form H2. So take the total from ammonia and divide by two to find the moles of H2 required.
the required equation is HgCl2+4KI>>2KCl+K2HgI4. according to stoichiometric calculations 4 moles of KI gives 1 mole of k2HgI4 THEREFORE 0.4 moles of K2HgI4 requires----- ? 0.4 moles x 4 moles/1 mole=1.6 moles therefore 1.6 moles of KI is required to produce 0.4 moles of K2HgI4
3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O is the equation if it is dilute nitric acid. In concentrated nitric acid the equation is different. So 3 moles of copper produce 2 moles of NO. Therefore it requires 6 moles of copper to produce 4 moles of NO.
This is not a common reaction at standard temperature and pressure.
800 g oxygen are needed.
N2 + 3H2 -> 2NH3 The stoichiometric equation (or balanced equation) for the formation of ammonia from this we can read off the mole ratio between hydrogen and ammonia; 3M H2 needed to produce 2M NH3 times each by 9 (so the ratio remains the same and 18M NH3 is formed) 27M H2 needed to produce 18M NH3
2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
This is based on calculations too. It contains 18 hydrogen moles.
Balanced equation first. 2CH4O + 3O2 -> 2CO2 + 4H2O 23.5 moles methanol (3 moles O2/2 mole CH4O) = 35.3 moles oxygen needed --------------------------------------
3,75 moles hydrogen
The answer would be 0.50 or .5
If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.
Assuming a decomposition reaction with this equation: 2KClO3(s) --> 2KCl(s) + 3O2(g), the ratio is 2:3, and if you produce 15mol O2, then 10mol potassium chlorate are needed.