For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
2HgO(s)�¨2Hg(l)+O 2 (g) 2\;{\rm{HgO}}\left( s \right)\; \rightarrow \;2\;{\rm{Hg}}\left( l \right) + {\rm{O}}_2 \left(1.40 kgHgO g \right)
I can't make sense of your post but here is how you work the problem.
1. Write and balanced equation. You've done that.
2. Convert grams HgO (whatever that is) to mols. mols = grams/molar mass
3. Using the coefficients in the balanced equation, convert mols HgO to mols O2.
4. Now convert mols O2 to grams O2. grams = mols x molar mass
To calculate how many grams of oxygen form when 1.40 kg of HgO completely reacts, we can use the molar ratio between HgO and O2 in the balanced equation.
The molar ratio between HgO and O2 is 2:1. This means that for every 2 moles of HgO, 1 mole of O2 is produced.
1 mole of HgO has a molar mass of 216.59 g (2 x 200.59 g/mol for Hg + 16.00 g/mol for O).
Therefore, 1.40 kg of HgO is equal to 1400 grams (1 kg = 1000 g), which is equal to 1400 g / 216.59 g/mol = 6.46 moles of HgO (rounded to two decimal places).
Using the molar ratio from the balanced equation, we can calculate the moles of O2 produced:
6.46 moles HgO x (1 mole O2 / 2 moles HgO) = 3.23 moles O2.
Finally, we need to convert moles of O2 to grams of O2. The molar mass of O2 is 32.00 g/mol.
3.23 moles O2 x 32.00 g/mol = 103.36 grams of O2.
Therefore, when 1.40 kg of HgO completely reacts, approximately 103.36 grams of O2 will form.
To calculate the number of grams of oxygen formed in the reaction, you'll need to use the molar ratio between mercury(II) oxide (HgO) and oxygen (O2). This ratio is determined from the balanced chemical equation.
In the balanced equation 2HgO(s) → 2Hg(l) + O2(g), the coefficient of HgO is 2 and the coefficient of O2 is also 2. This means that for every 2 moles of HgO that react, 2 moles of O2 are produced.
To calculate the number of moles of HgO, you need to divide the given mass by the molar mass of HgO. The molar mass of HgO can be obtained by adding the atomic masses of mercury (Hg) and oxygen (O) in the compound. The atomic masses are found on the periodic table.
The atomic mass of Hg is 200.59 g/mol, and the atomic mass of O is 16.00 g/mol. Therefore, the molar mass of HgO is (200.59 g/mol) + (16.00 g/mol) = 216.59 g/mol.
Now, we can calculate the number of moles of HgO. Given that the mass of HgO is 1.40 kg (notice that the mass needs to be converted to grams):
Mass of HgO = 1.40 kg x 1000 g/kg = 1400 g
Number of moles of HgO = Mass of HgO / Molar mass of HgO
= 1400 g / 216.59 g/mol
Now, using the molar ratio from the balanced equation, we can determine the number of moles of O2 formed:
Number of moles of O2 = Number of moles of HgO x (2 moles of O2 / 2 moles of HgO)
Finally, we can convert the number of moles of O2 to grams by multiplying by the molar mass of O2:
Mass of O2 = Number of moles of O2 x Molar mass of O2
Using this process, you can calculate how many grams of oxygen form when each quantity of reactant completely reacts.