Dentrification is the process of converting nitrate to nitrogen gas and nitrous oxide.
Answer:
In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.
Explanation:
Step 1: Data given
Mass of HCl = 69.0 grams
Mass of Al = 78.0 grams
Molar mass HCl = 36.46 g/mol
Atomic mass Al = 26.99 g/mol
Step 2: The balanced equation
6HCl(aq) + 2Al(s) → 3H2(g) + 2AlCl3(s)
Step 3: calculate moles HCl
Moles HCl = mass HCl / molar mass HCl
Moles HCl = 69.0 grams / 36.46 g/mol
Moles HCl = 1.89 moles
Step 4: Calculate moles Al
Moles Al = 78.0 grams / 26.99 g/mol
Moles Al = 2.89 moles
Step 5: Calculate the limiting reactant
For 6 moles HCl we need 2 moles Al to produce 3 moles H2 and 2 moles AlCl3
HCl is the limiting reactant. It will completely be consumed (1.89 moles). Aluminium is the reactant in excess. There will react 1.89 / 3 = 0.63 moles
There will remain 2.89 - 0.63 = 2.26 moles aluminium
In this reaction HCl is the limiting reactant and aluminium is the reactant in excess.
Answer:
Empirical formula is C₃H₄O
Explanation:
Empirical formula:
It is the simplest formula gives the ratio of atoms of different elements in small whole number
.
Given data:
Percentage of hydrogen = 7.14%
Percentage of carbon = 64.3%
Percentage of oxygen = 28.6%
Empirical formula = ?
Solution:
Number of gram atoms of H = 7.14 / 1.01 = 7.1
Number of gram atoms of O = 28.6 / 16 = 1.8
Number of gram atoms of C = 64.3 / 12 = 5.4
Atomic ratio:
C : H : O
5.4/1.8 : 7.1/1.8 : 1.8/1.8
3 : 4 : 1
C : H : O = 3 : 4 : 1
Empirical formula is C₃H₄O.
Answer:
D (hope I’m right, sorry if it’s not)