Answer:
172 g Al
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 26.98 101.96
4Al + 3O₂ ⟶ 2Al₂O₃
m/g: 325
(a) Calculate the <em>moles of Al₂O₃
</em>
n = 325 g Al₂O₃ × 1 mol Al₂O₃ /39.10 g Al₂O₃
n = 3.188 mol Al₂O₃
(b) Calculate the <em>moles of Al
</em>
The molar ratio is (4 mol Al/2 mol Al₂O₃)
n = 3.188 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃)
n = 6.375 mol Al
(c) Calculate the <em>mass of Al</em>
m = 6.375 mol Al × (26.98 g Al/1 mol Al)
m = 172 g Al
Note: The answer can have only <em>three</em> significant figures because that is all you gave for the mass of Al₂O₃.
Liquid silver is less dense than solid silver, so the solid silver would sink.
D. All of the answers are true
Your answer would be 172.1703 g/ml
we are given the the two reactants: AgNO3 and Na2CO3 and is asked to write a balanced equation and a net ionic equation for the reaction of the two. This is a double-replacement reaction:
2AgNO3 (aq)+ Na2CO3 (aq)= Ag2CO3 + 2NaNo3 (aq)
2 Ag + + 2 N03- + 2Na+ + CO32- = Ag2CO3 + 2 Na+ 2NO3-
cancelling the spectator ions, 2Ag + + CO32- = Ag2CO3
