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a) Gọi số mol Ca, CaCO3 là a, b (mol)
=> 40a + 100b = 19 (1)
\(m_{HCl}=\dfrac{500.4,38}{100}=21,9\left(g\right)\)
PTHH: Ca + 2HCl --> CaCl2 + H2
a--->2a------->a----->a
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b------>2b------>b------>b
=> \(\overline{M}_Y=\dfrac{2a+44b}{a+b}=13,6.2=27,2\left(g/mol\right)\)
=> 25,2a = 16,8b (2)
(1)(2) => a = 0,1 (mol); b = 0,15 (mol)
\(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)
b)
mdd sau pư = 19 + 500 - 0,1.2 - 0,15.44 = 512,2 (g)
mHCl(dư) = 21,9 - 36,5(2a + 2b) = 3,65 (g)
mCaCl2 = 111(a + b) = 27,75 (g)
\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{27,75}{512,2}.100\%=5,418\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{512,2}.100\%=0,713\%\end{matrix}\right.\)
a) Gọi số mol Ca, CaCO3 là a, b (mol)
=> 40a + 100b = 19 (1)
\(m_{HCl}=\dfrac{500.4,38}{100}=21,9\left(g\right)\)
PTHH: Ca + 2HCl --> CaCl2 + H2
a--->2a------->a----->a
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b------>2b------>b------>b
=> \(\overline{M}_Y=\dfrac{2a+44b}{a+b}=13,6.2=27,2\left(g/mol\right)\)
=> 25,2a = 16,8b (2)
(1)(2) => a = 0,1 (mol); b = 0,15 (mol)
\(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)
b)
mdd sau pư = 19 + 500 - 0,1.2 - 0,15.44 = 512,2 (g)
mHCl(dư) = 21,9 - 36,5(2a + 2b) = 3,65 (g)
mCaCl2 = 111(a + b) = 27,75 (g)
\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{27,75}{512,2}.100\%=5,418\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{512,2}.100\%=0,713\%\end{matrix}\right.\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PTHH: \(n_{Mg}=n_{H_2}=0,6\left(mol\right)\)
=> \(m_{Mg}=0,6.24=14,4\left(g\right)\)
=> \(m_{Cu}=50-14,4=35,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Mg=\dfrac{14,4}{50}.100=28,8\%\\\%Cu=\dfrac{35,6}{50}.100=71,2\%\end{matrix}\right.\)
\(n\)H2 =\(\dfrac{13,44}{22,4}\) =0,6(mol)
PTHH:
Mg +HCl →MgCl2 + H2
0,6 mol ←0,6 mol
a) \(m\)Mg =0,6. 24 =14,4(g)
\(m\)Cu= 50- 14,4= 35,6(g)
b)\(m\)%Mg= \(\dfrac{14,4}{50}\).100%= 28,8%
\(m\)%Cu=100%- 28,8%= 71,2%
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
Bạn xem lại xem đề cho bao nhiêu gam hỗn hợp nhé, vì mZn đã bằng 13 (g) rồi.
BaCO3 + 2HCl -> BaCl2 + CO2 + H2O (1)
MgSO3 + 2HCl -> MgCl2 + SO2 + H2O (2)
\(\overline{M}\)=24,5.2=49
Đặt nCO2=nBaCO3=a
nSO2=nMgSO3=b
\(\dfrac{44a+64b}{a+b}=49\)
=>\(\dfrac{a}{b}=3\)
Ta có hệ:
\(\left\{{}\begin{matrix}197a+104b=34,75\\a=3b\end{matrix}\right.\)
=>a=0,15;b=0,05
mBaCO3=197.0,15=29,55(g)
mMgSO3=104.0,05=5,2(g)
hỗn hợp khí sau pứ gồm khí HCl, CO2