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Oduvanchick [21]
4 years ago
11

Write the formula equation for the reaction between hydroiodic acid and beryllium hydroxide

Chemistry
1 answer:
Aleksandr-060686 [28]4 years ago
8 0
 The  formula  equation for  the reaction  between  hydroiodic  acid  and  beryllium  hydroxide      is as below

Be(OH)2  +  2 HI   =  BeI2 +   2H2O


Formula  of  hydroiodic is  =  HI
formula   of  beryllium hydroxide = Be(OH)2


1 mole  of  Be(OH)2  react  with   2  mole  of HI to form  1 mole  of BeI2  and   2   mole  of H2O
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a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

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Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)

To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:

6.5 g × (1mol / 315.48g) =<em> 0.0206moles of  Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:

0.0206moles of  Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:

0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>

8 0
4 years ago
Read 2 more answers
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