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klio [65]
3 years ago
11

Chemistry help on a Lab I did ?

Chemistry
1 answer:
lisov135 [29]3 years ago
8 0
<span>Well #        # Drops of Iron Sulfate             # Drops of Sodium Hydroxide
</span><span>  A1         5     45
  A2        10    40
  A3        15     35
  A4        20     30
  A5        25     25
  A6        30     20
  B1        35     15
  B2        40     10
  B3        45      5
</span>
<span>that's for B, because it says "Iron Sulfate".
which is this equation: FeSO4 (aq) + 2NaOH (aq)  Fe(OH)2 (s) + Na2SO4 (aq)
</span>
<span>FeSO4 (aq) + 2NaOH (aq)  Fe(OH)2 (s) + Na2SO4 (aq)
ratio of FeSO4:NaOH is 1:2,
so you need to moles of NaOH for every mole of FeSO4.
</span>
<span>A1-A6 had precipitate, though less ppt every time. This corresponds to the decrease in the number of NaOH drops At A2 10/1:40/2 the stoichiometric ratio is equal, so you should've seen the post ppt present. But looking at the picture it doesn't look right. I think you might've labelled them wrong, or there is an error somewhere. This fits micro-plate A better.</span><span>
</span>
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1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n
Ede4ka [16]

Answer:

m_{H2SO4 = 61.25 g

m_{Na2SO4} = 88.75 g

Explanation:

m_{NaOH} = \frac{200 . 25 }{100} = 50 g

⇒ n_{NaOH} = \frac{50}{40} = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

4 0
3 years ago
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