Đáp án + Giải thích các bước giải:

$a)m_{CaCO_3}=500.90\%=450(g)$

$CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow$

Bảo toàn KL:

$m_{CaCO_3}=m_{CR}+m_{CO_2}$

$\to m_{CO_2}=450-311,9=138,1(g)$

$\to n_{CO_2}=\dfrac{138,1}{44}=\dfrac{1381}{440}(mol)$

Theo PT: $n_{CaCO_3(p/ứ)}=n_{CO_2}=\dfrac{1381}{440}(mol)$

$\to H=\dfrac{\dfrac{1381}{440}.100}{450}.100\%\approx 69,75\%$

$b)$ Theo PT: $n_{CaO}=\dfrac{1381}{440}(mol)$

$\to m_{CaO}=\dfrac{1381}{440}.56\approx 175,76(g)$

$\to m_{CaCO_3(dư)}=450-\dfrac{1381}{440}.100\approx 136,14(g)$

$\to \%m_{CaO}=\dfrac{175,76}{175,76+136,14}.100\%\approx 56,35\%$

Đáp án:

Giải thích các bước giải:

$m_{CaCO_3}=500.90\%=450(g)$

Bảo toàn khối lượng:

$m_{CO_2}=m_{CaCO_3}-m_{R}=450-311,9=138,1(g)$

$n_{CO_2}=\dfrac{138,1}{44}=\dfrac{1381}{440}(mol)$

$CaCO_3\xrightarrow{t^o}CaO+CO_2$

Theo pt: $n_{CaCO_3}=n_{CO_2}=\dfrac{1381}{440}(mol)$

$\to H=\dfrac{\dfrac{1381}{440}.100}{500}.100\%=69,74\%$

$b)$

$n_{CaO}=n_{CO_2}=\dfrac{1381}{440}(mol)$

$m_{CaO}=\dfrac{1381}{440}.56=175,76(g)$

$\to \%m_{CaO}=\dfrac{175,76}{311,9}.100\%=56,35\%$