Excess reactant : Na
NaCl produced : = 16.497 g
<h3>Further explanation</h3>
Given
Reaction(balanced)
2Na + Cl₂⇒ 2NaCl
20 g Na
10 g Cl₂
Required
Excess reactant
NaCl produced
Solution
mol Na(Ar = 23 g/mol) :
= 20 : 23 = 0.87
mol Cl₂(MW=71 g/mol):
= 10 : 71 g/mol = 0.141
mol : coefficient :
Na = 0.87 : 2 = 0.435
Cl₂ = 0.141 : 1 = 0.141
Limiting reactant : Cl₂(smaller ratio)
Excess reactant : Na
Mol NaCl based on mol Cl₂, so mol NaCl :
= 2/1 x mol Cl₂
= 2/1 x 0.141
= 0.282
Mass NaCl :
= 0.282 x 58.5 g/mol
= 16.497 g
Answer:
Percentage of first isotope = 69.152 %
Percentage of second isotope = 30.848 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
For first isotope:
Let % = x %
Mass = 62.9296 amu
For second isotope:
% = 100 - x % (Since, there are only two isotopes)
Mass = 64.9278 amu
Average mass = 63.546 amu
Thus,
Solving,
1.9982 x = 138.18
Thus,
<u>Percentage of first isotope = x = 69.152 %</u>
<u>Percentage of second isotope = 100 - x % = 30.848 %</u>
Hydrogen peroxide decomposes to yield water and oxygen gas
That is; H2O2 (l) = H2O (l) + O2(g)
The standard heat of formation; H2O2 (l) = -187.6 kJ/mol; H2O(l) = -285.8 kJ/mol
1 mole of hydrogen peroxide contains 34 g
Thus, 5.4 g contains 5.4/34 = 0.1588 moles
The moles of water produced will also be equivalent to 0.1588 moles
Heat = heat of formation of product - reactant
Therefore; Heat = (0.1588 moles × -285,8 )- (0.1588× -187.6)
= -15.594 kJ
