It’s b for sure that’s the answer
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.
Balanced Equation: CuCl + NaNO₃ → NaCl + CuNO₃
Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol
the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,
then moles of NaCl = 0.31 mol
Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g
⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.
Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %
NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.
An aqueous solution of an arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt.
<h3>What is a Salt?</h3>
This is a compound which is formed as a result of a neutralization reaction between acid and base.
Arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt due to increased concentration of H+ and OH- respectively.
Read more about Salt here brainly.com/question/13818836
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Answer:
Chlorine is the element located in period 3, group 17 of the periodic table.
Explanation:
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
