There are four Hydrogen atoms in one molecule of Methane (CH₄).
And there are 6.022×10²³ molecules in 1 mole of CH₄.
So,
Number of Hydrogen atoms is 1 mole of CH₄ are,
= 6.022 × 10²³ × 4
= 2.4 ×10²⁴ Hydrogen Atoms
Now calculating for 2 moles,
As,
1 mole of CH₄ contains = 2.4 ×10²⁴ Hydrogen Atom
Then,
2 moles of CH₄ will contain = X Hydrogen Atoms
Solving for X,
X = (2 moles × 2.4 ×10²⁴ Hydrogen Atom) ÷ 1 mole
X = 4.8 × 10²⁴ Hydrogen Atoms
Answer is: 18 moles of lead(II)-nitrate.
Balanced chemical reaction:
3Pb(NO₃)₂ + 2AlCl₃ → 2Al(NO₃)₃ + 3PbCl₂.
n(Al(NO₃)₃) = 12 mol.
From chemical reaction: nAl(NO₃)₃) : n(Pb(NO₃)₂) = 2 : 3.
n(Pb(NO₃)₂) = 3 · 12 mol ÷ 2.
n(Pb(NO₃)₂) = 18 mol; amount of substance.
Al(NO₃)₃ is aluminium nitrate.
AlCl₃ is aluminium chloride.
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The answer to your question would be as followed:
The most precise measurement for the cylinder you are mentioning is <em><u>B. 43.0mL</u></em>
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Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol