The periods inside the Periodic Table is B. a horizontal row, and is numbered 1 through 7
hope this helps
Answer:
44.8 L of O2 will react (option D)
Explanation:
Step 1: Data given
Number of moles of SO2 = 4.00 moles
STP = Pressure = 1 atm and temperature = 273 K
Step 2: The balanced equation
2 SO2(g) + O2(g) → 2 SO3(g)
Step 3: Calculate moles of O2
For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3
For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2
Step 4: Calculate volume of O2
For 1 mol we have a volume of 22.4 L
V = (n*R*T)/ p
V = (2.00 * 0.08206 * 273)/p
V = 44.8 L
For 2.00 moles we have a volume of 2*22.4 = 44.8 L
44.8 L of O2 will react (option D)
WHAT THE HECK!?!?!?! AM I SUPPOSE TO KNOW WHATEVER LANGUAGE THAT IS?!?!
Answer:
This is a typical stoichiometry question.To answer this question you want to get a relationship between
N
a
2
O and NaOH.
So you can get a relationship between the moles of
N
a
2
O
and moles of NaOH by the concept of stoichiometry.
N
a
2
O +
H
2
O ----------------> 2 NaOH.
According to above balanced equation we can have the stoichiometry relationship between
N
a
2
O and NaOH. as 1:2
It means 1 moles of
N
a
2
O is required to react with one mol of
H
2
O to produce 2 moles of NaOH.
in terms of mass 1 mole of
N
a
2
O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.
62 g of
N
a
2
O produces 80 g of NaOH.
1g of NaOH is produced from 62/80 g of
N
a
2
O
1.6 x
10
2
g of NaOH will require 62 x 1.6 x
10
2
g / 80 of
N
a
2
O
124g of
N
a
2
O.
Explanation:
