Paul is conducting an experiment where he has combined 52.0 grams of CaO with 10

Question

Paul is conducting an experiment where he has combined 52.0 grams of CaO with 100.0 mL of water at 25°C, releasing some amount of steam as well. This produces the following equation:

Paul is conducting an experiment where he has combined 52.0 grams of CaO with 100.0 mL of water at 25 releasing some amount of steam as well. This produces the following equation: CaO(s) +H2O(l) Ca(OH),(s) a) Which of the two reactants are in excess and how many grams of this reactant will remain after the reaction is complete? Number O Cao b.) Using the information provided in the table below: SubstanceA(kJ/mol) CaO(s) H20(?) Ca(OH2(s 986.09 -635.1 - 285.84 Assuming all heat is transferred to the water, what is the mass of steam that was released? Number

Explanation / Answer

Solution-

Given-

Mass of CaO =52.0 g

Volume of H2O = 100.0 mL

Mass of H2O = d * V = 1 g mL-1* 100 mL = 100 g                 ( density of H2O = 1)

# mole of CaO = Mass/Molar mas = 52.0 g/56.07 g/mol = 0.927 mol of CaO

# mole of H2O = Mass/Molar mas = 100/18 = 5.55 mol of H2O

a) Which of two reactant are in excess & how many gram of reactant remain after the reaction is compete?
CaO(s) + H2O(l) Ca(OH)2(s)

In the above reaction 1 mole of CaO require 1 mole of H2O

Here limiting factor is 0.927 mol

Total mole of water =5.555 – 0.927 = 4.62 mol of water excess

Mass of water remaining after completion of reaction = # of remaining mol * molar mass = 4.62 mol * 18 g mol-1 = 83.3 g

Answer- Only Water is in excess in the reaction. 83.3 g of water remains after completion of reaction.

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