Answer:
290.82g
Explanation:
The equation for the reaction is given below:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol
Mass of H2SO4 from the balanced equation = 3 x 98 = 294g
Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96] = 54 + 288 = 342g
Now, we can obtain the mass of aluminium sulphate formed by doing the following:
From the equation above:
294g of H2SO4 produced 342g of Al2(SO4)3.
Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3
Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.
Noble gas notation for molybdenum:
[Kr] 4d^5 5s^1
Combination
Hope this helps :)
It actually depends on the percentage of the concentration give. Percentages can be expressed as %mass/mass, %volume/volume or %mass/volume. To keep things simple, let's just assume that it is in %volume/volume. Thus, 13% of 520 mL is pure acid.
Volume of pure acid = 520*0.13 = 67.6 mL
