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alisha [4.7K]
3 years ago
6

Please help, I'm struggling with stoichiometry

Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

a) HNO3

b) 26.8g (3 s.f.)

c) 1.29g (3 s.f.)

Please see the attached pictures for full solution.

To balance an equation, ensure that the number of atoms for each element is the same on both sides.

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2 years ago
what is the percent yield of NaCl if 31.0 g of CuCl reacts with excess NaNo3 to produce 21.2 g of NaCl
olchik [2.2K]
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.
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