There would be about 1.67 x 10^25 oxygen atoms and about 3.34 x 10^25 hydrogen atoms.
Answer:
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of compound = 0.519 g
Mass of CO₂ = 1.24 g
Mass of H₂O = 0.255 g
Empirical formula = ?
Solution:
%age of C,H,O
C = 1.24 g/0.519 × 12/44 ×100 = 65.5%
H = 0.255 g/0.519 × 2.016/18 ×100 = 5.6%
O = 100 - (65.5+5.6)
O = 28.9%
Number of gram atoms of H = 5.6 / 1.01 = 5.5
Number of gram atoms of O = 28.9 / 16 = 1.81
Number of gram atoms of C = 65.5 / 12 = 5.5
Atomic ratio:
C : H : O
5.5/1.81 : 5.5/1.81 : 1.81/1.81
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.
The mass of NaOH is 45 g
The molar mass of NaOH = 40 g/mol
The moles of NaOH = mass / molar mass
= 45 / 40
= 1.125
Thus, 1.125 mol NaOH is available
3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl
3 mol of NaOH react with 1 mol of FeCl₃
1.125 moles of NaOH will react with x moles of FeCl₃
x = 1.125 / 3
x = 0.375 mol
0.375 mol FeCl₃ can take part in reaction
The molar mass of FeCl₃ is 162 g/mol
The mass of FeCl₃ = moles × mass
= 0.375 × 162
= 60.75 g
Thus, the amount of FeCl₃, which can be consumed is 60.75 g
Answer : The rate of consumption of 
is, 
Explanation : Given,
Moles of = 
Volume of solution = 170 mL = 0.170 L (1 L = 1000 mL)
Time = 170 s
First we have to calculate the concentration.
Now we have to calculate the rate of consumption.
Thus, the rate of consumption of is,
