Calculate the mass of water having same number of hydrogen atoms as are present in the 32gram of methane. Please help me soon...
1 answer:
You might be interested in
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
The question is incomplete, here is the complete question:
A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.
<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
We are given:
Given mass of liquid bromine = 766. g
Molar mass of liquid bromine, = 159.8 g/mol
Putting values in above equation, we get:
Hence, the amount of liquid bromine produced is 4.79 moles.