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
Well I’m. Or to sure but it can’t be B because when you throw the ball the the kinetic energy is still increasing
The molarity of the hydrochloric acid, HCl solution given the data is 0.079 M
<h3>Balanced equation </h3>
HCl + NaOH —> NaCl + H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the molarity of HCl </h3>
- Molarity of base, NaOH (Mb) = 0.137 M
- Volume of base, NaOH (Vb) = 16.1 mL
- Volume of acid, HCl (Va) = 27.9 mL
- Molarity of acid, HCl (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 27.9) / (0.137 × 16.1) = 1
(Ma × 27.9) / 2.2057 = 1
Cross multiply
Ma × 27.9 = 2.2057
Divide both side by 27.9
Ma = 2.2057 / 27.9
Ma = 0.079 M
Thus, the molarity of the HCl solution is 0.079 M
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