Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
1 1 1 1
0,3 0,3 0,3 0,3
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
a). \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒\(V_{H2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
b). \(80ml=0,08l\)
\(n_{H2SO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
→\(C_M=\dfrac{n}{V}=\dfrac{0,3}{0,08}=3,75\left(M\right)\)
c). \(n_{MgSO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{MgSO4}=n.22,4=0,3.22,4=6,72\left(l\right)\)
→\(C_M=\dfrac{n}{V}=\dfrac{0,3}{6,72}=0,04\left(M\right)\)
d). \(MgSO_4+Ba\left(OH\right)_2\rightarrow Mg\left(OH\right)_2+BaSO_4\downarrow\)
1 1 1 1
0,3 0,3 0,3
\(n_{BaSO4\uparrow}=\dfrac{0,3.1}{1}\)=0,3(mol)
→\(m_{BaSO4\downarrow}=n.M=0,3.233=69,9\left(g\right)\)
\(n_{Ba\left(OH\right)_2}=\dfrac{0,3.1}{1}\)=0,3(mol)
\(\rightarrow V_{ddBa\left(OH\right)_2}=\dfrac{n}{C_M}=\dfrac{0,3}{1,6}=0,1875\left(l\right)\)
\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có: \(n_{CH_3COONa}=\dfrac{9,84}{82}=0,12\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=n_{NaOH}=n_{CH_3COONa}=0,12\left(mol\right)\)
\(\Rightarrow V_{ddCH_3COOH}=\dfrac{0,12}{0,5}=0,24\left(l\right)\)
\(m_{NaOH}=0,12.40=4,8\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{4,8}{20\%}=24\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,2 0,4 0,2
\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6.100\%}{20\%}=73\left(g\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(25^oC,1bar\right)}=0,2.22,4=4,48\left(l\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(n_{Na2CO3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH3OONa+CO_2+H_2O\)
0,2 0,1 0,2 0,1
a) \(V_{CO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
\(C_{MCH3COOH}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
b) \(C_{MCH3COONa}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Chúc bạn học tốt
a)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15-->0,3----->0,15-->0,15
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
c) \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
=> \(C\%=\dfrac{10,95}{200}.100\%=5,475\%\)
\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)
\(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\)
\(0.1..........0.1................0.1...........0.1\)
\(C_{M_{H_2SO_4}}=\dfrac{0.1}{0.2}=0.5\left(M\right)\)
\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{CaSO_4}=0.1\cdot136=13.6\left(g\right)\)
