If 33.5 g of Mg(OH)2 reacted, what mass of H2O could be produced? Mg(OH)2(s) + HCl(aq) ----> MgCl2(aq) + H2O(l)
I have balanced the equation for you.
Mg(OH)2(s) + 2HCl(aq) ----> MgCl2(aq) + H2O(l)
mols Mg(OH)2 in 33.5 g is g/molar mass = 33.5/58.3 = estimated 0.6 but you need to confirm all of these numbers.
For every 1 mol Mg(OH)2 you get 1 mol H2O so estimated 0.6 mols H2O.
Then grams H2O = mols H2O x molar mass H2O = ?
Post your work if you get stuck.
To determine the mass of water (H2O) that could be produced when 33.5 g of Mg(OH)2 reacts, you need to calculate the stoichiometric ratio between Mg(OH)2 and H2O in the balanced equation.
First, write down the balanced equation for the reaction:
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
According to the balanced equation, for every 1 mole of Mg(OH)2 that reacts, 2 moles of H2O are produced.
Next, calculate the number of moles of Mg(OH)2 using its molar mass. The molar mass of Mg(OH)2 is calculated by adding up the atomic masses of each element:
Mg (Magnesium) = 24.31 g/mol
O (Oxygen) = 16.00 g/mol (two in Mg(OH)2)
H (Hydrogen) = 1.008 g/mol (two in Mg(OH)2)
Molar mass of Mg(OH)2 = 24.31 g/mol + 16.00 g/mol + 1.008 g/mol + 1.008 g/mol = 41.328 g/mol
Now, divide the given mass of Mg(OH)2 (33.5 g) by its molar mass to calculate the number of moles:
Number of moles of Mg(OH)2 = Mass / Molar mass
Number of moles of Mg(OH)2 = 33.5 g / 41.328 g/mol
Finally, multiply the number of moles of Mg(OH)2 by the stoichiometric ratio of H2O to Mg(OH)2 to find the number of moles of water, and then multiply by the molar mass of H2O (18.015 g/mol) to find the mass:
Mass of H2O = (Number of moles of Mg(OH)2) x (2 moles of H2O / 1 mole of Mg(OH)2) x (18.015 g/mol)
Basically, multiply the moles of Mg(OH)2 by the ratio of moles of H2O to moles of Mg(OH)2, and then multiply by the molar mass of H2O.
By following these calculations, you can determine the mass of water (H2O) that could be produced when 33.5 g of Mg(OH)2 reacts.
To determine the mass of H2O that could be produced, we need to identify the stoichiometry of the reaction. From the balanced equation:
1 mol Mg(OH)2 reacts to produce 1 mol H2O
Now, let's calculate the molar mass of Mg(OH)2:
Molar mass of Mg(OH)2 = (1 × molar mass of Mg) + (2 × molar mass of O) + (2 × molar mass of H)
= (1 × 24.31 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol)
= 24.31 g/mol + 32.00 g/mol + 2.02 g/mol
= 58.33 g/mol
Next, we can determine the number of moles of Mg(OH)2 that reacted:
n(moles) = mass ÷ molar mass
= 33.5 g ÷ 58.33 g/mol
≈ 0.575 mol
According to the stoichiometry of the reaction, for every 1 mol of Mg(OH)2, 1 mol of H2O is produced.
Therefore, the mass of H2O that could be produced is equal to the moles of Mg(OH)2 that reacted:
mass of H2O = n(moles) × molar mass of H2O
= 0.575 mol × molar mass of H2O
Since we don't have the molar mass of H2O specified, let's assume it is approximately 18.02 g/mol (molar mass of water).
mass of H2O = 0.575 mol × 18.02 g/mol
≈ 10.37 g
Therefore, approximately 10.37 g of H2O could be produced when 33.5 g of Mg(OH)2 reacts.