Answer:
4.6 × 10²³ molecules:
Step-by-step solution
You will need a balanced equation with masses, moles, and molar masses, so let's gather the information in one place:
M_r: 22.99
2Na + 2H₂O ⟶ 2NaOH + H₂
m/g: 35
1. Calculate the <em>moles of Na
</em>
Moles of Na = 35 g Na × (1 mol Na/22.99 g Na)
Moles of Na = 1.52 mol Na
2. Calculate the <em>moles of H₂
</em>
Moles of H₂ = 1.52 mol Na × (1 mol H₂/2 mol Na)
Moles of H₂= 0.761 mol H₂
3. Calculate the molecules of H₂
6.022 × 10²³ molecules H₂ = 1 mol H₂
Molecules of H₂ = 0.761 × (6.022 × 10²³
/1)
Molecules of H₂ = 4.6 × 10²³ molecules H₂
The reaction forms 4.6 × 10²³ molecules of H₂.
A chemical reaction between the metals and the electrolyte frees more electrons in one metal than it does in the other. The metal that frees more electrons develops a positive charge, and the other metal develops a negative charge.
The complete equation for this reaction is,
Ba(NO3)2 + Na2PO4 = 2NaNO3 + BaPO4
Among the compounds present in the reaction, Barium Nitrate, Sodium Phosphate and Sodium Nitrate are soluble ionic compounds. Hence, they will completely ionize into ions. Only BaSO4 is insoluble which becomes the precipitate. Ionic equation is:
Ba2+ + 2NO3- + 2Na+ + PO42- = 2Na+ +NO3- + BaPO4
Cancel like ions,
Ba2+ + PO42- = BaPO4.
Thus, the stoichiometric coefficient of Barium ion is 1.
1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3
2. Na₂CO₃ as a limiting reactant
<h3>Further explanation</h3>
Given
Reaction
2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃
Required
mol ratio
Limiting reactant
Solution
The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)
1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3
2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67
Na₂CO₃ as a limiting reactant (smaller)
Answer:
When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.
