2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
Explanation:
To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
Bellow we have the balanced chemical equation of the complete combustion of C₃H₇OH:
C₃H₇OH (l) + (9/2) O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)
to have integer coefficients we multiply the reaction with 2:
2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
where:
(l) - liquid
(g) - gaseous
Learn more about:
combustion reaction
brainly.com/question/9425444
balancing chemical equations
brainly.com/question/13941483
#learnwithBrainly
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Answer:
C) Highly reactive
Explanation:
An atom with one or two valence electrons more than a closed shell is highly reactive, because the extra valence electrons are easily removed to form a positive ion
Explanation:
The given data is as follows.
Concentration of solution = 0.5 M
Volume of solution = 1 L
Molar mass of Glycylglycine = 132.119 g/mol
As molarity is the number of moles present in liter of solvent.
Mathematically, Molarity = 
Hence, calculate the number of moles as follows.
No. of moles = Molarity × Volume
= 
= 0.5 mol
Therefore, mass of glycylglycine = mol × molar mass
= 
= 66.06 g
Thus, we can conclude that 66.06 g glycylglycine is required.
