Stoichiometry is the ratio of reactants to products in a balanced chemical reaction equation.
the stoichiometry in this case of H₂ to O₂ is 2:1.
This means that 2 moles of H₂ reacts with 1 mol of O₂.
we have been asked to calculate the volume of H₂ gas.
Since both O₂ and H₂ are gases, at standard conditions its stated that molar volume of gases is 22.4 L
This means that at standard temperature and pressure, 1 mol of any gas occupies a volume of 22.4 L
first we need to calculate the number of O₂ moles reacted;
22.4 L of gas - 1 mol of O₂
1 L of gas - 1/22.4 mol/L
then 20 L of O₂ - 1/22.4 mol/L * 20 L = 0.89 mol
stoichiometry of H₂ to O₂ is 2:1
the number of H₂ moles = 0.89*2 = 1.79 mol
1 mol occupies 22.4 L
Therefore 1.79 mol = 22.4 L/mol * 1.79 mol = 40 L
Therefore it can be seen that stoichiometry applies to volumes as well.
volume of H₂ : O₂ = 40 L : 20 L = 2:1
volume and moles both can be determined by stoichiometry.
Volume of H₂ reacted = 40 L
Answer:
<h3>5.06282 × 10²⁴ molecules</h3>
Explanation:
The number of molecules of Ca2(SO3) can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 8.41 × 6.02 × 10²³
We have the final answer as
<h3>5.06282 × 10²⁴ molecules</h3>
Hope this helps you
Answer:
5.76g
Explanation:
2H2 + O2 => 2H2O
Molar mass of H2O = 18g/Mol
No of mole of H2O = 6.42/18 = 0.36 moles of water.
1 mole of O2 gives 2 moles of water.
Xmole of O2 gives 0.36 moles of water.
X = 0.36/2 = 0.18 mole
Molar mass of O2 = 32g/mol
Mass = 32 × 0.18 = 5.76g.
<span>magnesium cations (Mg </span><span>2+</span>) and oxide anions (<span>O 2</span>−<span>)</span>
C) Lower,
They're lower because molecular solids form IMF's such as Dipole-Dipole and H-Bonds, while Ionic bonds are much more powerful.