Question

Answer 1

a)

$Cu + 2H_2SO_{4_{đặc}} \to CuSO_4 + SO_2 + 2H_2O$

$n_{SO_2} = n_{Cu} = \dfrac{6,4}{64} = 0,1(mol)$
$V = 0,1.22,4 = 2,24(lít)$

b)

$n_{CuSO_4} = n_{Cu} = 0,1(mol)$
$m_{CuSO_4} = 0,1.160 = 16(gam)$

c)

$n_{H_2SO_4} = 2n_{Cu} = 0,2(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,2.98}{90\%} = 21,78(gam)$

Answer 2

\(n_{Cu}=\dfrac{6.4}{64}=0.1\left(mol\right)\)

\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(0.1.......0.2...............0.1.............0.1\)

\(V_{SO_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{CuSO_4}=0.1\cdot160=16\left(g\right)\)

\(m_{dd_{CuSO_4}}=\dfrac{0.2\cdot98\cdot100}{90}=21.78\left(g\right)\)