Respuesta :
We have a balanced equation:
2 Al+ 3 CuSO4 ⇒ Al2(SO4)3+ 3 Cu
1.85 g Al* (1 mol Al/ 26.98 g Al)* (3 mol Cu/ 2 mol Al)* ( 63.55 g Cu/ 1 mol Cu)= 6.54 g Cu.
(Note that the units cancel out so you get the answer)
6.54 g Cu* (56.6/100)= 3.70 g Cu.
The final answer is 3.70 g Cu~
2 Al+ 3 CuSO4 ⇒ Al2(SO4)3+ 3 Cu
1.85 g Al* (1 mol Al/ 26.98 g Al)* (3 mol Cu/ 2 mol Al)* ( 63.55 g Cu/ 1 mol Cu)= 6.54 g Cu.
(Note that the units cancel out so you get the answer)
6.54 g Cu* (56.6/100)= 3.70 g Cu.
The final answer is 3.70 g Cu~
Molar mass of :
Al = 27 g/moles
Cu = 63.5 g/moles
Balanced chemical equation:
3 CuSO4 + 2 Al = 1 Al2(SO4)3 + 3 Cu
2* 27 g Al ------------------> 3 * 63.5 g Cu
1.85 g Al -------------------> x g Cu
54 x =352.425
x = 352.425 / 54
x = 6.5263 g Cu
6.5263 g --------------- 100%
y g ----------------------- 56.6%
y = 56.63 . 6.5263 / 100
y = 369.38858 /100
y = 3.70 g of Cu
Al = 27 g/moles
Cu = 63.5 g/moles
Balanced chemical equation:
3 CuSO4 + 2 Al = 1 Al2(SO4)3 + 3 Cu
2* 27 g Al ------------------> 3 * 63.5 g Cu
1.85 g Al -------------------> x g Cu
54 x =352.425
x = 352.425 / 54
x = 6.5263 g Cu
6.5263 g --------------- 100%
y g ----------------------- 56.6%
y = 56.63 . 6.5263 / 100
y = 369.38858 /100
y = 3.70 g of Cu