Number of moles are calculated using the equation n=m/M
where n is the number of moles, m is the mass of the substance and M is the molar mass of the substance.
Molar mass of water = 18 g mol-1
Number of moles of water = 3.8 g /18 g mol-1
= 0.21 mol
Number of moles of H in H2O =2
(the subscript next to the chemical symbol of an element indicates the number of moles of that element in that molecule)
Number of moles of H in 3.8 g of H2O = 2 x 0.21 mol
= 0.42 mol
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 → 2NH3 From the equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, when 1.2 moles of H2 react, we can calculate the moles of NH3 produced as: 1.2 mol H2 * (2 mol NH3 / 3 mol H2) = 0.8 mol NH3.
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
2H2 + O2 --> 2H2OAs you can see by the balanced reaction, for every 1 mole of oxygen used, 2 moles of water are formed. Also notice that for every 1 mole of oxygen used, you need 2 moles of hydrogen to produce the 2 moles of water. So in your case 110 moles of oxygen would produce 220 moles of water & would also require 220 moles of hydrogen (which you have in excess since you have 230 moles of hydrogen). So 220 moles of water are the most that can be formed.
In excess quantity of hydrogen (H2) 5.0 moles water can be formed maximally from 2.5 mole oxygen,because 2.5 x 2 (atoms O per mole oxygen, O2) = 5.0 x 1 (atoms O per mole water, H2O), thus balancing this reaction for oxygen atoms (5 on both sides).
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 → 2NH3 From the equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, when 1.2 moles of H2 react, we can calculate the moles of NH3 produced as: 1.2 mol H2 * (2 mol NH3 / 3 mol H2) = 0.8 mol NH3.
3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
The answer is o,5 mol.
moles of what?
You use the Mole-to-Mole ratio. If the equation is 2CH4 + 2H2O = 6H2 + 2CO, then you would start with your given, 8.0 mol CO and multiply that with your mol-to-mol ratio which is (2mol CO/ 2 mol CH4). Your answer will be 8.0 mol.
1 mole of sulfuric acid=98g(?)chemical formula of sulfuric acid=H2SO4Atomic mass of Hydrogen(H)=1Atomic mass of Sulpher(S)=32Atomic mass of Oxygen(O)=16(1)2+32+(16)4=2+32+64=98uTo calculate number of molesNo. of moles=given mass/atomic massno. of moles=49/98no. of moles is 0.5g
2H2 + O2 --> 2H2OAs you can see by the balanced reaction, for every 1 mole of oxygen used, 2 moles of water are formed. Also notice that for every 1 mole of oxygen used, you need 2 moles of hydrogen to produce the 2 moles of water. So in your case 110 moles of oxygen would produce 220 moles of water & would also require 220 moles of hydrogen (which you have in excess since you have 230 moles of hydrogen). So 220 moles of water are the most that can be formed.
In excess quantity of hydrogen (H2) 5.0 moles water can be formed maximally from 2.5 mole oxygen,because 2.5 x 2 (atoms O per mole oxygen, O2) = 5.0 x 1 (atoms O per mole water, H2O), thus balancing this reaction for oxygen atoms (5 on both sides).
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
To form ammonia, balanced reaction is N(2) + 3H(2) ---> 2NH(3) + H(2)O. As you can see for 1 mole of nitrogen three moles of hydrogen is required. Hence for your question, 3 moles nitrogen is required to satisfy the ratio.
The molar mass of O is 16 g/mol and H is 1 g/mol. From the given masses, we calculate moles: 24 g O / 16 g/mol = 1.5 mol O and 16 g H2 / (2 g/mol) = 8 mol H2. The balanced chemical equation for the formation of water from hydrogen and oxygen is 2H2 + O2 -> 2H2O, so we need twice as many moles of H2 as O2. Hence, 1.5 mol O2 would require 3 mol H2, which is 3/2 * (2 g H2O/mol) = 3 g H2O.