Question

Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide CaOg + H₂2 Ca(OH)2) In a particular experiment a 2.00g sample of Calis reacted with excess water and 2.14g of Ca (OH) is recovered What is the percent held in this experiment

Answer 1

Molar mass of CaO,

MM = 1*MM(Ca) + 1*MM(O)

= 1*40.08 + 1*16.0

= 56.08 g/mol

mass of CaO = 2 g

mol of CaO = (mass)/(molar mass)

= 2/56.08

= 3.566*10^-2 mol

According to balanced equation

mol of Ca(OH)2 formed = moles of CaO

= 3.566*10^-2 mol

Molar mass of Ca(OH)2,

MM = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

mass of Ca(OH)2 = number of mol * molar mass

= 3.566*10^-2*74.1

= 2.643 g

% yield = actual mass*100/theoretical mass

= 2.14*100/2.643

= 81.0 %

Answer: 81.0 %