Answer:
1. 59.1 g HgO
2. A.) 38.2 L
3. B.) It is directly proportional to the number of moles of the gas.
Step-by-step explanation:
1. Mass of HgO
We know we will need a chemical equation with masses and molar masses, so let's gather all the information in one place.
M_r: 216.59
2HgO ⟶ 2Hg + O₂
(a) Calculate the <em>moles of O₂</em>
pV = nRT Divide each side by RT
n = (pV)/(RT)
<em>Data:
</em>
p = 0.970 atm
V = 4.50 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 390.0 K
<em>Calculation:
</em>
n = (0.970 × 4.500)/(0.082 06 × 390.0)
n = 0.1364 mol O₂
(b) Calculate the <em>moles of HgO
</em>
The molar ratio is 1 mol O₂/2 mol HgO.
Moles of HgO = 0.1364 mol O₂ × (2 mol Hg/1 mol O₂)
Moles of HgO = 0.2728 mol HgO
(c) Calculate the <em>mass of HgO
</em>
Mass of HgO = 0.2728 mol HgO × (216.59 g HgO/1 mol HgO)
Mass of HgO = <em>59.1 g HgO</em>
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2. Volume of Hydrogen
Mg + 2HCl ⟶ MgCl₂ + H₂
n/mol: 3.00
(a) Calculate the <em>moles of H₂
</em>
The molar ratio is (1 mol H₂/2 mol HCl).
Moles of H₂ = 3.00 mol HCl × (1 mol H₂/2 mol HCl)
Moles of H₂ = 1.50 mol HCl
(b) Calculate the <em>volume of H₂
</em>
pV = nRT Divide both sides by p
V = (nRT)/p
<em>Data:
</em>
n = 1.50 mol
T = 298 K
p = 0.960 atm
<em>Calculation:
</em>
V = (1.5 × 0.082 06 × 298)/0.960
V = 38.2 L
3. Volume of gas
pV = nRT
V = nRT/p
If T and p are constant, (RT)/p = constant = k
V = kn
V ∝ n
The volume is directly proportional to the number of moles of the gas.
