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icang [17]
3 years ago
6

Which of the following conversion factors would be used to calculate the number of moles of Cl2 produced if 11 moles of HCl were

present in the following reaction:______
4HCl O2 --> 2Cl2 2H2O
Chemistry
1 answer:
MissTica3 years ago
6 0

<u>Answer:</u> The moles of chlorine gas produced is 5.5 moles

<u>Explanation:</u>

We are given:

Moles of HCl = 11 moles

For the given chemical reaction:

4HCl+O2\rightarrow 2Cl_2+2H_2O

By Stoichiometry of the reaction:

4 moles of HCl produces 2 moles of chlorine gas

So, 11 moles of HCl will be produced from \frac{2}{4}\times 11=5.5mol of chlorine gas

Hence, the moles of chlorine gas produced is 5.5 moles

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Calculate the molarity of a solution of barium hydroxide if 18.15 ml of it is required for the titration of a 20.00 ml sample of
inysia [295]
The balanced equation for the reaction between Ba(OH)₂ and HCl is as follows;
Ba(OH)₂ + 2HCl ---> BaCl₂ + 2H₂O
stoichiometry of Ba(OH)₂ to HCl is 1:2
the number of HCl moles that have reacted - 0.2452 mol/L x (20.00 x 10⁻³ L)
number of HCl moles reacted = 0.004904 mol
2 mol of HCl reacts with 1 mol of Ba(OH)₂
therefore 0.004904 mol of HCl reacts with - 1/2 x 0.004904 mol of Ba(OH)₂
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6 0
3 years ago
Considering the limiting reactant, what is the mass of zinc sulfide produced from 0.250 g of zinc and 0.750 g of sulfur? Zn(s)+S
DIA [1.3K]

Answer:

The mass of zinc sulfide produced is  M_{ZnS} =  0.76 \ g

Explanation:

From the question we are told that

   The mass of zinc is  m_z =  0.750 \ g

    The mass of sulfur is  m_s =  0.250 \ g

The molar mass   of  Zn_{(s)}  is a constant with value  65.39 g /mol

The molar mass of S_{(s)}  is a constant with value  32.01 g/mol

The molar mass of  ZnS_{(s)} is a constant with value 97.46  g/mol

The reaction is  

        Zn_{(s)} + S_{(s)}  ------> ZnS_{(s)}

   So from the reaction

       1 mole of  Zn_{(s)} react with 1 mole of  S_{(s)} to produce 1 mole of ZnS_{(s)}

This implies that

65.39 g /mol of  Zn_{(s)} react with 32.01 g/mol of  S_{(s)} to produce   97.46  g/mol  of ZnS_{(s)}

From the values given we can deduce that the limiting reactant is sulfur cause  of the smaller mass

 So  

    0.250 g of  Zn_{(s)} react with 0.250 of  S_{(s)} to produce x \  g of  ZnS_{(s)}

So

      x =  \frac{97.46 * 0.250}{32.01}

       x =  0.76 \ g

Thus the mass of the mass of zinc sulfide produced is

    M_{ZnS} =  0.76 \ g

 

     

7 0
4 years ago
Please help due in 5 min.
Lina20 [59]

Answer:

B) Hope this helps! :)

Explanation:

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