Đáp án:
Giải thích các bước giải:
Gọi $\left \{ {{n_{Na}=a(mol)} \atop {n_{Ba}=b(mol)}} \right.$
$n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$
$2Na+2H_2O→2NaOH+H_2↑$
$a:a:a:0,5a$
$Ba+2H_2O→Ba(OH)_2+H_2↑$
$b:2b:b:b$
$∑n_{H_2}=0,5a+b=0,2(1)$
ddB: $\left \{ {{NaOH:a(mol)} \atop {Ba(OH)_2:b(mol)}} \right.$
$NaOH+HNO_3→NaNO_3+H_2O$
$0,5a:0,5a:0,5a:0,5a$
$Ba(OH)_2+2HNO_3→Ba(NO_3)_2+2H_2O$
$0,5b:b:0,5b:b:$
$m_{muối}=85.0,5a+261.0,5b=42,5a+130,5b=21,55(2)$
$(1),\ (2)⇒\left \{ {{a=0,2(mol)} \atop {b=0,1(mol)}} \right.$
a)$V_{ddHNO_3}=\frac{0,5.0,2+0,1}2=0,1(l)$
b)$m_{hợp\ kim}=23.0,2+137.0,1=18,3(g)$
%$m_{Na}=\frac{23.0,2}{18,3}.100=25,137$%
%$m_{Ba}=\frac{137.0,1}{18,3}.100=74,863$%
Xin hay nhất!!!