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
Answer:
3.79 moles
Explanation:
To convert moles to gams of a substance we need to find the molar mass of the substance. For Ca(OH)₂ th molar mass is:
1Ca = 40.08g/mol
2O = 2*16g/mol = 32g/mol
2H = 2*1.01g/mol = 2.02g/mol
The molar mass is:
40.08g/mol + 32g/mol + 2.02g/mol = <em>74.1g/mol</em>
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And moles are:
281g * (1mol / 74.1g) =
<h3>3.79 moles</h3>

chlorine's atomic number and mass :
35,453 u

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Answer:
The mass fraction of ferric oxide in the original sample :
Explanation:
Mass of the mixture = 3.110 g
Mass of 
Mass of 
After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

Mass of mixture left after all the ferric oxide has reacted = 2.387 g
Mass of mixture left after all the ferric oxide has reacted = y

The mass fraction of ferric oxide in the original sample :
