What mass of calcium hydroxide is formed when 10.0g of calcium oxide reacts with 10.0g of water?

The mass of calcium hydroxide that is formed when 10.0 g of CaO reacts with  10.0 g of water is 13.024 grams

calculation

from the equation

CaO + H2O → Ca(OH)2,

 1 moles of CaO reacted with 1 moles of H2O to form 1 moles of Ca(OH)2

find the moles of each reactant

moles=mass/molar mass

moles of CaO= 10 g/56 g/mol=0.179 moles

moles of H2O = 10 g/18 g/mol 0.556  moles

CaO is the limiting reagent  therefore by use of mole ratio of CaO:Ca(OH)2 which is 1:1 moles of Ca(OH)2 is = 0.179 moles

mass= moles x molar mass

= 0.176 moles x  74 g/mol = 13.024  grams

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3