Ta có: $100ml=0,1(l)$

a,$nCaCO_3=10/100=0,1(mol)$

$nH_2SO_4=0,1.1,5=0,15(mol)$

$mddH_2SO_4=1,1.100=110(g)$

$CaCO_3+H_2SO_4→CaSO_4+H_2O+CO_2↑$

Vì $0,1/1<0,15/1$ nên $H_2SO_4$ dư 0,05 mol hay dư 0,05.98=4,9 g

$nCaSO_4=0,1(mol)$

$⇒mCaSO_4=0,1.136=13,6(g)$

$∑mdd=110+10-0,1.44=115,6(g)$

⇒%$CaSO_4=13,6.100/115,6≈11,76$%

%$H_2SO_4 dư=4,9.100/115,6≈4,24$%

b,$H_2SO_4+2NaOH→Na_2SO_4+2H_2O$

$⇒nNaOH cần=0,1(mol)$

$⇒V_{ddNaOH4M}=0,1/4=0,025(l)=25(ml)$

$⇒m_{ddNaOH4M}=25.1,2=30(g)$

Đáp án:

\(\begin{array}{l} a,\ C_{M_{CaSO_4}}=1\ M.\\ C_{M_{H_2SO_4}}\text{(dư)}=0,5\ M.\\ C\%_{CaSO_4}=11,76\%\\ C\%_{H_2SO_4}\text{(dư)}=4,24\%\\ b,\ m_{\text{dd NaOH}}=30\ g.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:CaCO_3+H_2SO_4\to CaSO_4+CO_2↑+H_2O\\ n_{CaCO_3}=\dfrac{10}{100}=0,1\ mol.\\ n_{H_2SO_4}=0,1\times 1,5=0,15\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,1}{1}<\dfrac{0,15}{1}\\ \Rightarrow H_2SO_4\ \text{dư.}\\ \Rightarrow n_{H_2SO_4}\text{(dư)}=0,15-0,1=0,05\ mol.\\ Theo\ pt:\ n_{CaSO_4}=n_{CO_2}=n_{CaCO_3}=0,1\ mol.\\ \Rightarrow C_{M_{CaSO_4}}=\dfrac{0,1}{0,1}=1\ M.\\ C_{M_{H_2SO_4}}\text{(dư)}=\dfrac{0,05}{0,1}=0,5\ M.\\ m_{\text{dd H$_2$SO$_4$}}=100\times 1,1=110\ g.\\ \Rightarrow m_{\text{dd spư}}=m_{CaCO_3}+m_{\text{dd H$_2$SO$_4$}}-m_{CO_2}=10+110-0,1\times 44=115,6\ g.\\ \Rightarrow C\%_{CaSO_4}=\dfrac{0,1\times 136}{115,6}\times 100\%=11,76\%\\ \Rightarrow C\%_{H_2SO_4}\text{(dư)}=\dfrac{0,05\times 98}{115,6}\times 100\%=4,24\%\\ b,\ PTHH:H_2SO_4+2NaOH\to Na_2SO_4+2H_2O\\ Theo\ pt:\ n_{NaOH}=2n_{H_2SO_4}=0,1\ mol.\\ \Rightarrow V_{NaOH}=\dfrac{0,1}{4}=0,025\ lít=25\ ml.\\ \Rightarrow m_{\text{dd NaOH}}=25\times 1,2=30\end{array}\)

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