How many grams of mercury are formed from 47.0 of HgO?
How many grams of oxygen are formed from 47.0 of HgO?
How many grams of HgO would you need to obtain 34.1 of O2?
To answer these questions, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction involving mercury oxide (HgO). The balanced equation is:
2 HgO -> 2 Hg + O2
Let's start with the first question: How many grams of mercury (Hg) are formed from 47.0 grams of HgO?
To solve this, we will use the molar mass of HgO and the stoichiometric ratio between HgO and Hg.
1. Calculate the molar mass of HgO:
Molar mass of HgO = atomic mass of Hg + atomic mass of O.
The atomic masses of Hg and O are found on the periodic table.
2. Convert the given mass (47.0 grams) of HgO to moles:
Moles of HgO = Given mass (in grams) / Molar mass of HgO.
3. Use the stoichiometric ratio from the balanced equation:
From the balanced equation, we know that 2 moles of HgO produce 2 moles of Hg.
So, the molar ratio is 1:1 (moles of HgO to moles of Hg).
4. Calculate the mass of Hg formed:
Mass of Hg = Moles of HgO * Molar mass of Hg.
Follow the same steps to answer the second question: How many grams of oxygen (O2) are formed from 47.0 grams of HgO? Here, you will use the molar mass of O2 and the stoichiometric ratio between HgO and O2.
For the third question: How many grams of HgO would you need to obtain 34.1 grams of O2? You will follow a similar approach, but in this case, you will use the molar mass of O2 and the stoichiometric ratio between HgO and O2 to calculate the moles of O2, and then use the stoichiometric ratio between HgO and O2 to find the moles and mass of HgO needed.
To answer these questions, we need to calculate the molar masses of HgO and O2.
1. Molar Mass of HgO (Mercury(II) oxide):
Molar mass of Hg = 200.59 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of HgO = 200.59 + 16.00 = 216.59 g/mol
2. Calculation for Mercury (Hg):
To find the grams of mercury formed from 47.0 g of HgO, we need to convert grams of HgO to moles of HgO and then use the stoichiometry from the balanced chemical equation.
Given:
Mass of HgO = 47.0 g
Molar mass of HgO = 216.59 g/mol
Formula:
Mass (g) = Moles (mol) * Molar mass (g/mol)
Calculations:
Moles of HgO = Mass of HgO / Molar mass of HgO
Moles of HgO = 47.0 g / 216.59 g/mol
Moles of HgO = 0.2170 mol
From the balanced chemical equation:
1 mol HgO produces 1 mol Hg
Therefore, the number of moles of mercury (Hg) formed is also 0.2170 mol.
Now, to find the grams of mercury formed, we use the formula again:
Mass of Hg = Moles of Hg * Molar mass of Hg
Mass of Hg = 0.2170 mol * 200.59 g/mol
Mass of Hg = 43.40 g
Therefore, 47.0 g of HgO will form 43.40 g of mercury.
3. Calculation for Oxygen (O2):
To find the grams of oxygen formed from 47.0 g of HgO, we again need to use the stoichiometry of the balanced chemical equation.
From the balanced chemical equation:
1 mol HgO produces 1 mol O2
Therefore, the number of moles of oxygen (O2) formed is also 0.2170 mol.
Now, to find the grams of oxygen formed, we use the formula:
Mass of O2 = Moles of O2 * Molar mass of O2
The molar mass of O2 is 32.00 g/mol because O2 is a diatomic molecule.
Mass of O2 = 0.2170 mol * 32.00 g/mol
Mass of O2 = 6.94 g
Therefore, 47.0 g of HgO will form 6.94 g of oxygen.
4. Calculation for HgO:
To find the grams of HgO needed to obtain 34.1 g of O2, we will use the stoichiometry of the balanced chemical equation.
Given:
Mass of O2 = 34.1 g
Molar mass of O2 = 32.00 g/mol
We will use the inverse of the stoichiometry ratio between HgO and O2 to find the moles of HgO:
From the balanced chemical equation:
1 mol HgO produces 1 mol O2
So, 1 mol O2 produces 1 mol HgO.
Therefore, the moles of HgO needed to obtain 34.1 g of O2 is also 34.1 mol.
To find the grams of HgO needed, we can use the formula:
Mass of HgO = Moles of HgO * Molar mass of HgO
Mass of HgO = 34.1 mol * 216.59 g/mol
Mass of HgO = 7387 g
Therefore, you would need 7387 g of HgO to obtain 34.1 g of O2.
All of these stoichiometry problems are done the same say.
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