Antifreeze is an additive in water-based liquid to lower down the freezing point of such liquid. It is used to make use of the colligative properties of solutions specifically freezing-point depression for cold climate and boiling-point elevation to allow higher coolant temperature.
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________
<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
</span>
<span>n(CO)= m(CO) / M(CO)
n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>
