Answer:
See Explanation
Explanation:
moles of NH₃ = 11.9g/17.03 g/mol = 0.699 mole
moles of CN₂OH₄ = 1/2(0.699) mole =0.349 mole
Theoretical yield of CN₂OH₄ = (0.349 mole)(60 g/mole) = 20.963 grams
%Yield = Actual Yield/Theoretical Yield x 100%
= 18.5g/20.963g x 100% = 88.25%
Vertical Column is a group
Horizontal is Period
Aliki earth metals is a group
Lithium- Neon is both
Based on electrons are both
Number of electron shells is both
The mass of a 0.513 mol of Al2O3 is 52.33g.
HOW TO CALCULATE MASS:
The mass of a substance can be calculated by multiplying the molar mass of the substance by its number of moles. That is;
mass of Al2O3 = no. of moles of Al2O3 × molar mass of Al2O3
According to this question, there are 0.513 moles of Al2O3.
Mass of Al2O3 = 0.513 × 102
Mass of Al2O3 = 52.33g
Therefore, the mass of a 0.513 mol of Al2O3 is 52.33g.
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Answer:
Explanation:
Given parameters:
Mass of aluminium oxide = 3.87g
Mass of water = 5.67g
Unknown:
Limiting reactant = ?
Solution:
The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.
Using the number of moles, we can ascertain the limiting reactants;
Al₂O₃ + 3H₂O → 2Al(OH)₃
Number of moles;
Number of moles = 
molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole
number of moles = 
= 0.04mole
molar mass of H₂O = 2(1) + 16 = 18g/mole
number of moles = 
= 0.32mole
From the reaction equation;
1 mole of Al₂O₃ reacted with 3 moles of H₂O
0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O
But we were given 0.32 mole of H₂O and this is in excess of amount required.
This shows that Al₂O₃ is the limiting reactant
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