When magnesium oxide (MgO) and water (H20)react, they produce magnesium hydroxide Mg(OH)2. Magnesium hydroxide is a medicine used to treat an upset stomach. The equation for this chemical change is shown below:
MgO + H20 --> Mg(OH)2
if 40.00 grams of magnesium oxide and 18.00 grams of water are combined in this reaction, what will be the mass of the product? Explain.
58.00
This is a limiting reagent problem and you know that because amounts are given for BOTH reactants. I worked a problem just a few minutes ago and here is a link. Just follow the same steps.
http://www.jiskha.com/display.cgi?id=1418356094
Blablabla..answer later
Well if you had them, you will get 58.00
so 58
or 58.00
Oh, it's time to play with chemicals! Let's see what happens when magnesium oxide (MgO) and water (H2O) decide to mingle. According to the equation, they form magnesium hydroxide, or Mg(OH)2. Now, you want to know the mass of the product, right?
To find the mass of the product, we need to determine the limiting reactant. In other words, which reactant is going to run out first and prevent the reaction from going any further?
Magnesium oxide (MgO) has a molar mass of 40.30 g/mol, and water (H2O) has a molar mass of 18.02 g/mol. So let's calculate the number of moles we have of each.
For MgO:
40.00 g = 40.00 g/mol (molar mass of MgO)
And for H2O:
18.00 g = 18.00 g/mol (molar mass of H2O)
Now, we can compare the number of moles of each reactant. The reaction requires one mole of MgO and one mole of H2O to produce one mole of Mg(OH)2. But since the molar ratio is 1:1, the reactant with the fewer moles will determine the maximum amount of product we can get.
In this case, since 40.00 g of MgO is equal to 1 mole (40.00 g/mol), and 18.00 g of H2O is equal to 1 mole (18.00 g/mol), we can tell that H2O is the limiting reactant because it gives us the lower number of moles.
So, the reaction will consume all the 18.00 grams of water, and we'll be left with 1 mole of Mg(OH)2. The molar mass of Mg(OH)2 is 58.33 g/mol.
Therefore, the mass of the product Mg(OH)2 will be 58.33 grams. Voila!
To find the mass of the product, magnesium hydroxide (Mg(OH)2), we need to determine the limiting reactant. The limiting reactant is the reactant that will be completely consumed first, thus determining the maximum amount of product that can be obtained.
To find the limiting reactant, we need to compare the number of moles of each reactant.
First, we determine the number of moles of magnesium oxide (MgO):
Mass of MgO = 40.00 g
Molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Number of moles of MgO = Mass of MgO / Molar mass of MgO
= 40.00 g / 40.31 g/mol
â 0.9918 mol
Next, we determine the number of moles of water (H2O):
Mass of H2O = 18.00 g
Molar mass of H2O = (1.01 g/mol * 2) + 16.00 g/mol = 18.02 g/mol
Number of moles of H2O = Mass of H2O / Molar mass of H2O
= 18.00 g / 18.02 g/mol
â 0.9994 mol
Since the moles of reactants are approximately equal, we can say that the molar ratio of MgO to H2O is 0.9918 mol: 0.9994 mol, or approximately 1:1.
From the balanced equation, we know that the molar ratio of MgO to Mg(OH)2 is 1:1 as well. Therefore, from stoichiometry, the number of moles of Mg(OH)2 formed will be the same as the number of moles of MgO.
Number of moles of Mg(OH)2 = Number of moles of MgO
= 0.9918 mol
Now, we calculate the mass of Mg(OH)2 using the molar mass of Mg(OH)2:
Molar mass of Mg(OH)2 = (24.31 g/mol + (1.01 g/mol * 2)) + (16.00 g/mol + (1.01 g/mol * 2))
= 58.33 g/mol
Mass of Mg(OH)2 = Number of moles of Mg(OH)2 * Molar mass of Mg(OH)2
= 0.9918 mol * 58.33 g/mol
â 57.85 g
Therefore, the mass of the product, Mg(OH)2, will be approximately 57.85 grams.