From the balanced equation it can be seen that it takes 2 moles KOH to react with each 1 mole of MgCl2. So, the answer is 2.
The stoichiometry of the reaction determines the amount of Mg(OH)2 produced. In this case, the balanced equation shows that 1 mole of MgCl2 reacts with 2 moles of KOH to produce 1 mole of Mg(OH)2. Therefore, if 3 moles of MgCl2 are added, 6 moles of KOH are needed to completely react with it and produce 3 moles of Mg(OH)2.
1 mole of MgCl2 requires 2 moles of KOH to react based on the balanced chemical equation provided.
Potassium hydroxide is the limiting agent.
MgCl2 would be the limiting reagent
MgCl2 + 2KOH ==> Mg(OH)2 + 2KCl1 mole MgCl2 reacts with 2 moles KOH 2 moles KOH x 56.1 g/mole = 112.2 g KOH = 100 g KOH (to 1 significant figure based on 1 mole)
The answer is one mole.
The stoichiometry of the reaction determines the amount of Mg(OH)2 produced. In this case, the balanced equation shows that 1 mole of MgCl2 reacts with 2 moles of KOH to produce 1 mole of Mg(OH)2. Therefore, if 3 moles of MgCl2 are added, 6 moles of KOH are needed to completely react with it and produce 3 moles of Mg(OH)2.
1 mole of MgCl2 requires 2 moles of KOH to react based on the balanced chemical equation provided.
Potassium hydroxide is the limiting agent.
MgCl2 would be the limiting reagent
MgCl2 + 2KOH ==> Mg(OH)2 + 2KCl1 mole MgCl2 reacts with 2 moles KOH 2 moles KOH x 56.1 g/mole = 112.2 g KOH = 100 g KOH (to 1 significant figure based on 1 mole)
1
MgCl2(aq) + 2KOH(aq) --> Mg(OH)2(s) + 2KCl(aq)It is the molar ratio in the equation. Every mole of magnesium chloride requires 2 moles of potassium hydroxide. Thus 3 moles would need 6 moles of alkali for complete reaction. We don't have that much, so potassium hydroxide is the limiting reactant and we can only use 2 moles of the magnesium chloride and produce 2 moles of magnesium hydroxide.
Potassium hydroxide is the limiting reagent.
Potassium hydroxide is the limiting reagent.
Potassium hydroxide is the limiting reagent.
If the chemical reaction is:2 HCl + Mg = MgCl2 + 2 H23143 moles of HCl produce 3143 moles of hydrogenThis volume at 0 0C is70 447,202litres.