<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>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>
<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>
It has to be intensive, which means it does not depend on the amount of the substance. Boiling point, melting point, and density are three intensive physical properties.