How many grams of copper will be replaced in 2 L of a 1.50- M CuSO4 solution if the latter is made to react with 27.0 g of aluminium ? ( Cu =63.5, Al =27.0 ) (3 CuSO 4+2 Al arrow Al 2( SO 4)3+3 Cu )

Question

How many grams of copper will be replaced in 2 L of a 1.50- M CuSO4 solution if the latter is made to react with 27.0 g of aluminium ? ( Cu =63.5, Al =27.0 ) (3 CuSO 4+2 Al arrow Al 2( SO 4)3+3 Cu ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(W1/W2)=(E1/E2) (W1/27)=(63.5/2) × (3/27) W1=63.5 × (3/2) × (27/27)=95.25 g

Q. How many grams of copper will be replaced in $2 L$ of a $1.50 - M C u S O_{4}$ solution if the latter is made to react with $27.0 g$ of aluminium ? $(C u = 63.5, A l = 27.0)$
$(3 C u S O_{4} + 2 A l \to A l_{2} (S O_{4})_{3} + 3 C u)$

$\frac{W _{1}}{W _{2}} = \frac{E _{1}}{E _{2}}$
$\frac{W _{1}}{27} = \frac{63.5}{2} \times \frac{3}{27}$
$W_{1} = 63.5 \times \frac{3}{2} \times \frac{27}{27} = 95.25 g$

Q. How many grams of copper will be replaced in 2L of a 1.50−MCuSO4​ solution if the latter is made to react with 27.0g of aluminium ? (Cu=63.5,Al=27.0) (3CuSO4​+2Al→Al2​(SO4​)3​+3Cu)

W2​W1​​=E2​E1​​ 27W1​​=263.5​×273​ W1​=63.5×23​×2727​=95.25g

Q. How many grams of copper will be replaced in $2 L$ of a $1.50 - M C u S O_{4}$ solution if the latter is made to react with $27.0 g$ of aluminium ? $(C u = 63.5, A l = 27.0)$
$(3 C u S O_{4} + 2 A l \to A l_{2} (S O_{4})_{3} + 3 C u)$

$\frac{W _{1}}{W _{2}} = \frac{E _{1}}{E _{2}}$
$\frac{W _{1}}{27} = \frac{63.5}{2} \times \frac{3}{27}$
$W_{1} = 63.5 \times \frac{3}{2} \times \frac{27}{27} = 95.25 g$