[Marco245]

Good morning,I'm french and I try to understand a this exercice:

1)Give the formula of the intermetallic compound which forms magnesium with antimony according to the melting diagram of the Mg-Sb system.

2)What is the composition of the solid phase which precipitates the first if the liquid alloy is allowed to cool to 60% by mass of antimony?

But I haven't understand how we know(we found) than we have S1+L,S2+L,B(s)+L etc...in the diagram?

And for the 1) My teacher says that 77% of Sb=nSb/nBtotal and 23% of Mg=nMg/nBtotal with B the compound Mg(p)Sb(q) and ultimatly my teacher says than q/p=2/3 so B is Mg3.Sb2.

But for the 2) I thought we had to use the graph and the 60% to reply.
 I'm lost,someone could help me please?
For 2)my teacher says :the compound is Mg3Sb2,it's all!

[mjc123]

1) If the compound is 77% by mass Sb, how many g of Sb are in 1 g of compound? How many moles of Sb is that?
Similarly, if it is 23% by mass Mg, how many g and how many moles of Mg in 1g?
So what is the mole ratio of Mg to Sb?

2) Start at the point (60%, 1000°C). Draw a vertical line to represent the alloy cooling at constant composition. Which region does this line first enter after leaving the one-phase liquid (L) region?

But I haven't understand how we know(we found) than we have S1+L,S2+L,B(s)+L etc...in the diagram?

I don't understand this? Do you mean you haven't studied phase diagrams? Then why has your teacher given you this question?

[Marco245]

Hello,for 1) we have 0,77g of Sb and 0.23g of Mg.
n(mole)=m/M=0.77/24,31=0.031 moles for Mg and for Sb 0.23/121.76=0.00188 moles.
0.031/0.00188=16.48.
2)no my teacher is good,I'm just a bad and not very intelligent student who want to understand this lesson.
I will try to translate well and understand well what you say to 2),but thank you very much for your help

[AWK]

Hello,for 1) we have 0,77g of Sb and 0.23g of Mg.
n(mole)=m/M=0.77/24,31=0.031 moles for Mg and for Sb 0.23/121.76=0.00188 moles.