$n_{B}$ = $\frac{1.008}{22,4}$ $= 0,045(mol)$
$n_{Ag}$ = $\frac{2,16}{108}$ $= 0,02 (mol)$
Giải thích các bước giải:
$CaCO_3 + 2HCl → CaCl_2 + CO_2 + H_2O$
$Mg + 2HCl → MgCl_2 + H_2$
$Cu + 2AgNO3 → Cu(NO_3)_2 + 2Ag$
Gọi $n_{CaCO_3} = x$
Gọi $n_{Mg} = y$
Gọi $n_{Cu} = z$
Ta có hpt
$$<=>\left\{ \begin{array}{ll} 100x + 24y + 64z = 4 \\ x + y = 0,045 \\ 2z = 0,02 \end{array} \right.$$ $$<=>\left\{ \begin{array}{ll} x = 0,03 \\ y = 0,015 \\ z = 0,01 \\ \end{array} \right.$$
%$m_{CaCO_3}$ = $\frac{0,03.100}{4}$ . 100% $= 75$%
%$m_{Mg}$ = $\frac{0,015.24}{4}$ . 100% $= 9$%
%$m_{Cu}$= $\frac{0,01.64}{4}$ . 100% $= 16$%
$n_{HCl}$ $= 2x + 2y = 2.(0,03+0,015) = 0,09 (mol)$
=> $m_{HCl}$ $= 0,09 . 36,5 = 3,285 (g)$
$m_{HCldd}$ $= 3,285 : 20$% $= 16,425 (g)$