the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment
theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations
we need to first find the theoretical yield
2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂
molar ratio of Na₂O₂ to Na₂CO₃ is 2:2
number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed
number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol
therefore number of Na₂CO₃ moles formed is - 0.10 mol
mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g
therefore theoretical yield is 10.6 g
percent yield = actual yield / theoretical yield x 100%
81.0 % = actual yield / 10.6 g x 100 %
actual yield = 10.6 x 0.81
actual yield = 8.59 g
therefore actual yield is 8.59 g
Answer:
B is correct answer because The enthalpy is negative so this means the reaction produces heat and the reaction is exothermic
Answer:
2Na⁺ (aq) and 2OH⁻(aq)
Explanation:
Spectator ions:
Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.
Given ionic equation:
Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)
In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq) net ionic equation can be written as,
Net ionic equation:
Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)
A. 1
b. 1
c. 1
d. 4
As u can see H is the only one with 4 atoms as in the chemical formula it has a little 4 next to it