How many grams of Mg(OH)2 would be produced from 4 mol KOH? if the balanced reaction is MgCl2 +2KOH arrow Mg(OH)2 + 2KCl?
note that each 2 moles of KOH produce 1 mole of Mg(OH)2.
So, how many grams in 2 moles of magnesium hydroxide?
To determine the number of grams of Mg(OH)2 produced from 4 mol of KOH, we need to use stoichiometry, which involves the ratios between the reactants and products in a balanced chemical equation.
In this case, the balanced chemical equation is:
MgCl2 + 2KOH → Mg(OH)2 + 2KCl
From the equation, we can see that 1 mol of MgCl2 reacts with 2 mol of KOH to produce 1 mol of Mg(OH)2. This means that the stoichiometric ratio between MgCl2 and Mg(OH)2 is 1:1.
Given that we have 4 mol of KOH, we can determine the amount of Mg(OH)2 produced as follows:
4 mol KOH × (1 mol Mg(OH)2 / 2 mol KOH) = 2 mol Mg(OH)2
Now, to convert moles to grams, we need to know the molar mass of Mg(OH)2.
The molar mass of Mg is approximately 24.31 g/mol, O is 16.00 g/mol, and H is 1.01 g/mol. Since there are two oxygen atoms and two hydrogen atoms in Mg(OH)2, the molar mass of Mg(OH)2 can be calculated as follows:
Mg: 1 × 24.31 g/mol = 24.31 g/mol
O: 2 × 16.00 g/mol = 32.00 g/mol
H: 2 × 1.01 g/mol = 2.02 g/mol
Total molar mass of Mg(OH)2: 24.31 g/mol + 32.00 g/mol + 2.02 g/mol = 58.33 g/mol
Finally, we can calculate the mass of Mg(OH)2 produced using the molar mass:
Mass of Mg(OH)2 = 2 mol × 58.33 g/mol ≈ 116.66 grams
Therefore, approximately 116.66 grams of Mg(OH)2 would be produced from 4 mol of KOH.