First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
___AlBr3 + ___K -> ___KBr + ___ Al
1 AlBr3 + 3K -> 3KBr + 1 Al
hope this helps............
I'm taking this lesson now, so imma help u ( if u need anything else ask me)
so given Molar mass= 32 g/mol
molar mass= (empirical formula) n
32 = (14x1 + 2x1) n
32 = 16 n , so n= 2
so, molecular formula= N2H4
Answer:
A. Whatever is in the water, it moves. Even if a block is placed in there, it would move by sinking to the bottom. If a plastic bag was placed in there, it wouldn't sink but move a little.
Answer:
The amount of NaOH required to prepare a solution of 2.5N NaOH.
The molecular mass of NaOH is 40.0g/mol.
Explanation:
Since,
NaOH has only one replaceable -OH group.
So, its acidity is one.
Hence,
The molecular mass of NaOH =its equivalent mass
Normality formula can be written as:
Substitute the given values in this formula to get the mass of NaOH required.
Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g
