If 0.605 g of magnesium hydroxide reacts with 0.900g of sulfuric acid, what is the mass of magnesium sulfate produced?

Mg(OH)2(s)+H2SO4(l)→MgSO4(s)+H2O(l)

same as last one I just did

To find the mass of magnesium sulfate produced in the reaction, we can follow these steps:

Step 1: Find the molar mass of each compound involved in the reaction.
- The molar mass of Mg(OH)2 (magnesium hydroxide) is calculated as follows:
Mg: 24.31 g/mol
O: 16.00 g/mol (there are two O atoms)
H: 1.01 g/mol (there are two H atoms)
Therefore, molar mass of Mg(OH)2 = (24.31 g/mol) + (2 * 16.00 g/mol) + (2 * 1.01 g/mol) = 58.31 g/mol

- The molar mass of H2SO4 (sulfuric acid) is calculated as follows:
H: 1.01 g/mol (there are two H atoms)
S: 32.07 g/mol
O: 16.00 g/mol (there are four O atoms)
Therefore, molar mass of H2SO4 = (2 * 1.01 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol) = 98.09 g/mol

- The molar mass of MgSO4 (magnesium sulfate) is calculated as follows:
Mg: 24.31 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (there are four O atoms)
Therefore, molar mass of MgSO4 = (24.31 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol) = 120.38 g/mol

Step 2: Determine the limiting reactant.
The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. To find the limiting reactant, we need to compare the number of moles of each reactant.

- Moles of Mg(OH)2 = mass/Molar mass = 0.605 g / 58.31 g/mol ≈ 0.0104 mol
- Moles of H2SO4 = mass/Molar mass = 0.900 g / 98.09 g/mol ≈ 0.0092 mol

Since the stoichiometric ratio between Mg(OH)2 and H2SO4 is 1:1, we can see that H2SO4 has fewer moles than Mg(OH)2. Therefore, H2SO4 is the limiting reactant.

Step 3: Calculate the mass of MgSO4 produced.
Since the stoichiometric ratio between Mg(OH)2 and MgSO4 is also 1:1, we can conclude that the moles of MgSO4 produced will be equal to the moles of H2SO4 used.

- Moles of MgSO4 produced = moles of H2SO4 used ≈ 0.0092 mol

Now, to find the mass of MgSO4, we can use the formula:

Mass = Moles * Molar mass

Mass of MgSO4 = 0.0092 mol * 120.38 g/mol ≈ 1.10 g

Therefore, the mass of magnesium sulfate produced in the reaction is approximately 1.10 grams.