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Rainbow [258]
3 years ago
7

How come the balanced equation for calcium oxide reaction with water is: CaO (s) +H2O (l) --> Ca(OH)2 (s)?

Chemistry
2 answers:
sladkih [1.3K]3 years ago
5 0
CaO (s) +H2O (l) --> Ca(OH)2 (s) 

<span># of moles of CaO and H2O </span>
<span>n = given mass / molar mass. </span>
<span>Ca = 40 </span>
<span>O = 16 </span>
<span>Mass CaO = 56 grams/mole </span>

<span>n = 2.50 / 56
grams/mole = 0.0446 moles </span>

<span>Molar mass = 2*H + O = 2 + 16 = 18 grams/ mol </span>

<span>n = 2.41 / 18 = 0.134 (WATER) </span>

Now,
<span>The number of moles of water left over. </span>

<span>moles of water - moles of water used in the reaction = left over moles. </span>
<span>0.134 - 0.0446 = 0.0893 
</span>
<span>The grams of water left over. </span>

<span>1 mole of water = 18 grams </span>
<span>0.0893 moles of water = X</span>

<span>X = 18 x 0.0893 = 1.61 grams of water not used in the reaction.</span>
maxonik [38]3 years ago
3 0

<span>CaO(s) + H2O(l) ---> Ca(OH)2(aq) 
</span><span>CaO(s) + H2O(l) ---> 2CaOH(aq) we get this after balancing this equation
</span><span>2CaO(s) + H2O(l) ---> 2CaOH(aq) 
</span><span> number of moles
</span>CaO 

<span>n = given mass / molar mass. </span>
<span>Ca = 40 </span>
<span>O = 16 </span>
<span>Mass CaO = 56 grams/mole </span>

<span>n = 2.50 / 56 grams/mole = 0.0446 moles 
</span>Molar mass = 2*H + O = 2 + 16 = 18 grams/ mol 
now in water
<span>n = 2.41 / 18 = 0.134 
</span>moles of water - moles of water used in the reaction = left over moles. 

<span>0.134 - 0.0446 = 0.0893 remaining number of moles

</span><span> grams of water which are left </span>

<span>1 mole of water = 18 grams </span>
<span>0.0893 moles of water = x </span>

<span>x = 18 * 0.0893 = 1.61 are grams of water used</span>
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Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

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Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

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Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

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