Respuesta :
13.5 x [tex]10^-2[/tex] g of aluminium will be produced if 1.0 g of Ba reacts with 1.80 g of [tex]Al_2(SO_4)3[/tex].
What are moles?
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilograms of carbon 12; its symbol is “mol”.
Balanced Equation:
[tex]3Ba + Al_2(SO_4)_3[/tex] → [tex]2Al+3BaSO_4[/tex]
Grams of Ba: 1.0 g
Grams of [tex]Al_2(SO_4)_3[/tex]: 1.8g
Calculate the moles of Ba and [tex]Al_2(SO_4)_3[/tex]:
[tex]\frac{1g Ba}{137.3}[/tex] = 7.3 x[tex]10^{-3}[/tex]mol
1.8g [tex]Al_2(SO_4)_3[/tex]/ 342 = 5.3 x [tex]10^-3[/tex] mol [tex]Al_2(SO_4)_3[/tex]
Find the limiting reactant:
Ba has a coefficient of 3 in the balanced equation, so we divide the moles of Ba by 3 to get 7.3 x [tex]10^-3[/tex] mol Ba ÷3 = 2.43 x [tex]10^-3[/tex]
[tex]Al_2(SO_4)_3[/tex] has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 x[tex]10^-3[/tex]
2.43 x [tex]10^-3[/tex] is smaller than 5.3 x [tex]10^-3[/tex], therefore Ba is the limiting reactant.
finally, we just find the number of moles of Al
The ratio of Al to Ba is 2:3
7.3 x [tex]10^-3[/tex] x (2÷3) = 5 x [tex]10^-3[/tex] mol Al
Convert into mass.
5 x [tex]10^-3[/tex] mol Al x 27 = 13.5 x [tex]10^-2[/tex] g
Hence, 13.5 x [tex]10^-2[/tex] g of aluminium will be produced if 1.0 g of Ba reacts with 1.80 g of [tex]Al_2(SO_4)3[/tex].
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