If 0.605 g of magnesium hydroxide reacts with 1.00g of sulfuric acid, what is the mass of magnesium sulfate produced?
Mg(OH)2(s)+H2SO4(l)→MgSO4(s)+H2O(l)
.605g Mg x 1molMgSO/1molMg=.605g MgSo
1.00g MgSO/1mol H2SO4=1.00g MgSO
.605g+1.00g=1.605g
This is incorrect, but I don't understand what else to do.
molar mass of Mg(OH)2 = 24.3+2*17 = 58.3 g
so we have
.605/58.3 = .0104 mol
molar mass of H2SO4 = 2+32+64 = 98 g
so we have
1/98 = .0102 mol
so
we will get .0102 mol of MgSO4
.0102 ( 24.3+32+64) = 1.23 g
Well, it seems like you're stuck in a bit of a chemical predicament. Don't worry, I'm here to help... and maybe throw in a little humor along the way!
To solve this problem, we need to use stoichiometry. So let's take a closer look at the balanced chemical equation:
Mg(OH)2(s) + H2SO4(l) → MgSO4(s) + H2O(l)
We know that in order for the reaction to occur, the reactants need to be in the same molar ratio as shown in the equation. In this case, for every 1 mole of Mg(OH)2, we need 1 mole of H2SO4 to react and produce 1 mole of MgSO4.
Now let's do some calculations:
Given:
Mass of Mg(OH)2 = 0.605g
Mass of H2SO4 = 1.00g
Step 1: Convert masses to moles
Moles of Mg(OH)2 = 0.605g / molar mass of Mg(OH)2
Moles of H2SO4 = 1.00g / molar mass of H2SO4
Step 2: Determine the limiting reactant
To find the limiting reactant, we compare the moles of each reactant and see which one is in excess. The reactant with the lesser number of moles is the limiting reactant.
Now, step back for a moment and think about it. Why did the scarecrow win an award? Because he was outstanding in his field! Brain teaser intermission!
Okay, back to business. Where were we? Ah yes, the limiting reactant. Once you figure out which one it is, you can move on to the next step.
Step 3: Calculate the moles of the product
Since we know the balanced chemical equation, we can use the mole ratio to calculate the moles of the product formed. In this case, for every 1 mole of MgSO4 produced, 1 mole of Mg(OH)2 reacts.
Now that we have the moles of the product, we can move on to the final step.
Step 4: Convert moles to mass
To calculate the mass of the product, simply multiply the moles of the product by its molar mass.
Now, give me a moment to crunch the numbers and I'll be right back with the answer!
To solve this problem, you need to determine the limiting reagent and then calculate the mass of magnesium sulfate produced.
1. Calculate the number of moles of each reactant:
- Moles of magnesium hydroxide (Mg(OH)2):
Molar mass of Mg(OH)2 = 24.31 g/mol (Mg) + 2 * 1.01 g/mol (H) + 2 * 16.00 g/mol (O) = 58.33 g/mol
moles of Mg(OH)2 = 0.605 g / 58.33 g/mol ≈ 0.01040 mol
- Moles of sulfuric acid (H2SO4):
Molar mass of H2SO4 = 2 * 1.01 g/mol (H) + 32.07 g/mol (S) + 4 * 16.00 g/mol (O) = 98.09 g/mol
moles of H2SO4 = 1.00 g / 98.09 g/mol ≈ 0.01019 mol
2. Determine the limiting reagent:
To find the limiting reagent, compare the ratio of moles between Mg(OH)2 and H2SO4. The stoichiometry of the balanced equation is 1:1, so the reagent with fewer moles is the limiting reagent. In this case, H2SO4 has fewer moles.
3. Calculate the moles of MgSO4 produced:
moles of MgSO4 = 0.01019 mol (moles of H2SO4, which is the limiting reagent)
4. Calculate the mass of MgSO4 produced:
Molar mass of MgSO4 = 24.31 g/mol (Mg) + 32.07 g/mol (S) + 4 * 16.00 g/mol (O) = 120.37 g/mol
mass of MgSO4 = 0.01019 mol × 120.37 g/mol ≈ 1.23 g
Therefore, the mass of magnesium sulfate produced is approximately 1.23 g.
To solve this problem, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, we need to convert the mass of each reactant into moles. To do this, we divide the given mass by the molar mass of the compound.
Given:
Mass of Mg(OH)2 = 0.605g
Molar mass of Mg(OH)2 = 58.32g/mol
0.605g Mg(OH)2 x (1 mol Mg(OH)2 / 58.32g Mg(OH)2) = 0.0104 mol Mg(OH)2
Given:
Mass of H2SO4 = 1.00g
Molar mass of H2SO4 = 98.09g/mol
1.00g H2SO4 x (1 mol H2SO4 / 98.09g H2SO4) = 0.0102 mol H2SO4
To determine the limiting reactant, we compare the mole ratios of the reactants as given by the balanced equation.
From the balanced equation, we can see that 1 mol of Mg(OH)2 reacts with 1 mol of H2SO4 to produce 1 mol of MgSO4.
The mole ratio of Mg(OH)2 to H2SO4 is 1:1.
Since the mole ratio is the same for both reactants, we can compare the moles directly.
0.0104 mol Mg(OH)2 vs 0.0102 mol H2SO4
Based on these calculations, we can see that the number of moles of Mg(OH)2 is slightly higher than the number of moles of H2SO4. Therefore, Mg(OH)2 is in excess and H2SO4 is the limiting reactant.
Now, let's calculate the mass of MgSO4 produced using the limiting reactant.
Using the mole ratio from the balanced equation, we know that 1 mol of H2SO4 will produce 1 mol of MgSO4.
So, the mass of MgSO4 produced can be calculated as:
0.0102 mol H2SO4 x (120.37g MgSO4 / 1 mol H2SO4) = 1.23g MgSO4
Therefore, the mass of MgSO4 produced is 1.23g.