Molar mass problem When a certain amount of liquid compound XY2 is completely burned in a certain amount of oxygen, the equation is XY2 (L) + 3O2 (g) = xo2 (g) + 2yo2 (g). After cooling, the volume of the product is 672 ml and the density is 2.56 g / L under the standard condition?

Molar mass problem When a certain amount of liquid compound XY2 is completely burned in a certain amount of oxygen, the equation is XY2 (L) + 3O2 (g) = xo2 (g) + 2yo2 (g). After cooling, the volume of the product is 672 ml and the density is 2.56 g / L under the standard condition?

XY2 (liquid) + 3O2 (gas) = = = = = xo2 (gas) + 2yo2 (gas) △ v = 0 (so the volume of gas after reaction is equal to that before reaction)
So V (O2) = 672ml = 0.672l
So n (O2) = 0.672l/22.4l/mol = 0.03mol
So m (O2) = 0.03mol * 32g / mol = 0.96g
XY2 (liquid) ---- 3O2 (gas)
1---------------------3
x---------------------0.03mol
So x = 0.01mol, that is, n [XY2 (liquid)] = 0.01mol
The volume of the product is 0.672 L, the density is 2.56 g / L, the mass of the product is 0.672 L * 2.56 g / L = 1.72 G
The mass of the product is 1.72g and the oxygen consumed is 0.96g, so m [XY2 (liquid)] = 1.72g-0.96g = 0.76g (according to the mass conservation),
M [XY2 (liquid)] = 0.76g, n [XY2 (liquid)] = 0.01mol, so m [XY2 (liquid)] = m / N = 76g / mol
10. Y is C and s, respectively

How many moles of hydrogen are produced per mole of aluminum
In the reaction of aluminum with aluminum hydrochloride and sodium hydroxide, the valence of aluminum varies from 0 to + 3, so how many moles of hydrogen are produced in the reaction of no mole of aluminum? The answer is 3 / 2. I want to know why

2Al+2NaOH+2H2O=2NaAlO2+3H2
2 3
1 X
x=3/2