Answer:
Option B, HCO3 1-
Explanation:
The valence of Sodium ion is +1 and the valence of HCO3 is -1. Thus, sodium ion has an extra electron to be donated to complete its outer shell while HCO3 needs an electron to complete its outer shell
Hence Na will combine with HCO3 to form NaHCO3
Option B is correct
Answer:
1.17 mol
Explanation:
Step 1: Write the balanced equation
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
Step 2: Calculate the moles corresponding to 85.0 g of HCl
The molar mass of HCl is 36.46 g/mol.
85.0 g × 1 mol/36.46 g = 2.33 mol
Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl
The molar ratio of HCl to H₂ is 6:3.
2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂
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:
B.
I think this is right.
If it is not I apologize