Magnesium and iron are metallic elements. How does a mole of magnesium compare with a mole of iron?
2 answers:
Answer: d) They both have the same number of atoms.
Explanation:
A mole is defined as the amount of substance that contains Avogardro number of the substance. Avogadro's number is given by .
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number
Thus 1 mole of magnesium and iron both contains particles.
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Answer:
1) 2.0 g
2) 0 g
3) 4.17 g
4) 2.57 g
Explanation:
First of all, we need to know the compounds and the reaction. The ion carbonate is , and the ion nitrate is
.
Sodium is in group 1, so it must lose one electron to be stable, and be the cation . Silver has only one electron too, so the cation will be
.
To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:
Sodium carbonate: Na2CO3
Silver nitrate: AgNO3
Silver carbonate: Ag2CO3
Sodium nitrate: NaNO3
The balanced reaction will be:
Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3
Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)
The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:
Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol
So the molar mass of the compounds must be:
Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)
AgNO3 = 170 g/mol (108 + 14 + 3x16)
Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)
NaNO3 = 85 g/mol
We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:
1 mol of Na2CO3 ---------- 2 mol of AgNO3
106 g ------------------------------ 2x170 = 340 g
x ------------------------------------ 5.14 g
By a simple direct three rule:
340x = 544.84
x = 1.6 g of Na2CO3
That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.
1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:
m = 3.6 - 1. 6 = 2.0 g
2) All the AgNO3 reacted, so there isn't a mass present after the reaction.
m = 0 g
3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3
2 moles of AgNO3 ------------- 1 mol of Ag2CO3
2x170 g ------------------------------- 276 g
5.14 g --------------------------------- x
By a simple direct three rule:
340x = 1418.64
x = 4.17 g of Ag2CO3
4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3
2 moles of AgNO3 ----------------------- 2 moles of NaNO3
2x170 g ---------------------------------------- 2x85 g
5.14 g ------------------------------------------- x
By a simple direct three rule:
340x = 873.8
x = 2.57 g