What is the sum of the coefficients (including “1”) of the balanced equation?
__CaCO3 + __NaF __ CaF2 + __Na2CO3
4
5
6
7
8
If you start with 20.0 grams CaCO3, how many moles of CaCO3 do you have?
20.0
2.00
.200
.0200
.0386
If you start with .500 moles of NaF, how many moles of CaF2 will form
.500
1.00
1.50
2.00
0.250
What is the sum of the coefficients (including “1”) of the following reaction?
Sodium sulfate + calcium hydroxide ???
4
5
6
7
9
Predict the precipitate that forms: HCl + AgNO 3 ???
HNO3
HCl
AgNO3
AgCl
no precipitate forms
The balanced reaction is
CaCO3 + 2 NaF -> CaF2 + Na2CO3
You forgot the reaction arrow.
The sum of the coefficients is 5.
We will be glad to critique your work on the remainder of the problem. I assume you can compute molar mass of a compound and convert mass to moles.
number 2 is .2 because I found molar mass. Number 3 4 and 5 is where I get lost.
I posted under my friends name accidentally.. but number 2 is .2 because I found molar mass. Number 3 4 and 5 is where I get lost
0.2 is correct for #2.
Here is how you work #3 and all problem similar to it. First, the copy.
CaCO3 + 2NaF ==> CaF2 + Na2CO3
If you start with .500 moles of NaF, how many moles of CaF2 will form
.500
1.00
1.50
2.00
0.250
ASSUMING you have an unlimited supply of CaCO3 (not the 20.0 g in problem 2), convert 0.500 mols NaF to mols CaF2 by using the coefficients in the balanced equation.
0.500 mols NaF x (1 mol CaF2/2 mol NaF) = (notice how the units of NaF cancel and mols CaF2 are left--this is the way we convert mols of one thing to mols of another) = 0.500 NaF x 1/2 = 0.250 CaF2.
On the others, if you get stuck, repost as a new post (at the top of the board) and tell us exactly what your trouble is in detail. That way we can help you better.
To find the sum of the coefficients (including "1") of a balanced equation, you need to count the number in front of each compound or element. In the given equation:
CaCO3 + NaF -> CaF2 + Na2CO3
The coefficients are:
CaCO3: 1
NaF: 2
CaF2: 1
Na2CO3: 1
The sum of these coefficients is: 1 + 2 + 1 + 1 = 5
Therefore, the sum of the coefficients is 5.
To calculate the moles of a substance, you need to use its molar mass.
In the second question, you are given that you start with 20.0 grams of CaCO3 and you need to find the number of moles.
The molar mass of CaCO3 is calculated by adding the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms:
CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol) = 100.09 g/mol
To find the number of moles, divide the mass (in grams) by the molar mass:
Number of moles = mass / molar mass = 20.0 g / 100.09 g/mol = 0.1999 ≈ 0.200 moles
Therefore, if you start with 20.0 grams of CaCO3, you have approximately 0.200 moles of CaCO3.
In the third question, you are given that you start with 0.500 moles of NaF and need to find the moles of CaF2 that will form.
From the balanced equation:
CaCO3 + NaF -> CaF2 + Na2CO3
The stoichiometry ratio between NaF and CaF2 is 1:1. This means that for every 1 mole of NaF, 1 mole of CaF2 will form.
Therefore, if you start with 0.500 moles of NaF, you will have 0.500 moles of CaF2.
To find the sum of the coefficients for the given reaction:
Sodium sulfate + calcium hydroxide -> ???
You need to balance the equation first. Balancing the equation ensures that the number of atoms on both sides of the reaction remains the same.
For the given equation, Na2SO4 + Ca(OH)2 -> ???
Balancing the equation gives:
Na2SO4 + Ca(OH)2 -> Na2SO4 + 2Ca(OH)2
The coefficients are:
Na2SO4: 1
Ca(OH)2: 2
The sum of these coefficients is: 1 + 2 = 3
Therefore, the sum of the coefficients is 3.
To predict the precipitate that forms in the reaction:
HCl + AgNO3 -> ???
You need to identify the product that would be insoluble in water and form a solid precipitate.
In this case, HCl is a strong acid and AgNO3 is a soluble salt. When a strong acid reacts with a soluble salt, a double displacement reaction occurs resulting in the formation of a solid precipitate.
In this case, the precipitate that would form is AgCl (silver chloride) because it is insoluble in water.
Therefore, the precipitate that forms in the reaction is AgCl.