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jonny [76]
3 years ago
8

0 ml of a 1.20 m solution is diluted to a total volume of 228 ml. a 114-ml portion of that solution is diluted by adding 111 ml

of water. what is the final concentration? assume the volumes are additive.
Chemistry
1 answer:
ozzi3 years ago
3 0
54.0 ml of a 1.2 m solution was diluted to a total volume of 228 ml
USing dilution equation
M1V1 = M2V2
Where M1V1 is before dilution while M2V2 is after dilution.
Therefore;
M1= 1.20 M, V2= 54 ml,  
M2=              V2 = 228 ml

1.20 M × 54 ml = M2 × 228 ml
  M2 = (1.2 ×54)/ 228 ml
       = 0.2842 M

0.2842 × 114 ml = M2 × (114 +111)
      M2 = (0.2842 ×114)/ 225
            = 0.143995
            = 0.144 M

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How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl
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Answer:

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

Explanation:

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Mg + 2 HCl → MgCl₂ + H₂

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Being:

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  • H: 1 g/mole
  • Cl: 35.45 g/mole

the molar mass of the compounds participating in the reaction is:

  • Mg: 24.31 g/mole
  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
  • MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
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Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

  • Mg: 1 mole* 24.31 g/mole= 24.31 g
  • HCl: 2 moles* 36.45 g/mole= 72.9 g
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  • H₂: 1 mole* 2 g/mole= 2 g

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}

moles of H₂= 0.134

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