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Arte-miy333 [17]
3 years ago
13

What is the approximate percent yield for this reaction? Ba(NO3)2 + Na2SO4 —> BaSO4 + 2 NaNO3

Chemistry
2 answers:
melomori [17]3 years ago
6 0

The percent yield for this reaction is 91%.

<u>Explanation</u>:

                   Ba(NO3)2 + Na2SO4 — > BaSO4 + 2NaNO3  

  • Assume that 0.45 mol of Ba(NO3)2 reacts with excess Na2SO4 to create 0.41 mol of BaSO4.  
  • Convert to grams of BaSO4 utilizing its molar mass (233.38 g/mol), which will be giving the actual yield.  
  • Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4/1 mol BaSO4)  

                               = 96 g BaSO4  

  • Beginning with 0.45 mol of Ba(NO3)2, compute the theoretical yield of BaSO4. Mole proportion of Ba(NO3)2 to BaSO4 is 1:1.  
  • 0.45 mol Ba(NO3)2 x (1 mol BaSO4/1 mol Ba(NO3)2) x (233.38 g BaSO4/1 mol BaSO4) = 105 g BaSO4  

                    % Yield = (Actual Yield/Theoretical Yield) x 100%  

                              = (96/105) x 100%  

                               = 91%.

pantera1 [17]3 years ago
4 0

Answer:

Is (B)

Explanation:

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If 4.88 grams of zn react with 5.03 grams of s8 to produce 6.02 grams of zns, what are the theoretical yield and percent yield o
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<span>Balance the equation
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    work out how much ZnS is possible if all of each reagent were to react. The reagent that produces the least product is the limiting reagent.
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    moles S8 = 5.03 g / 256.48 g/mol = 0.01961 mol
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    The amount of Zn provided gives the least ZnS, therefore Zn is the limiting reagent.
    The theoretical yield is the maximum possible yield possible from the reagents provided. It is the amount of product that will form if all the limiting reagent reacts.
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If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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