a) \(n_{MgO}=\frac{6}{40}=0,15mol\)
PTHH : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
Theo PTHH: \(n_{H_2SO_4}=n_{MgO}=0,15mol\)
=> \(m_{H_2SO_4}=0,15.98=14,7g\)
b) \(m_{dd}=1,2.50=60g\)
\(C\%=\frac{14,7.100}{60}=24,5\%\)
c)Theo PTHH : \(n_{MgSO_4}=n_{MgO}=0,15mol\Rightarrow m_{MgSO_4}=0,15.120=18g\)
\(m\)dd sau p/ứng = 6 +60 = 66 g
C% sau p/ứng = \(\frac{18.100}{66}=27,\left(27\right)\%\)
\(n_{MgO}=\frac{m}{M}=\frac{6}{40}=0,15\left(mol\right)\)
\(PTHH:MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
(mol) 1 1 1 1
(mol) 0,15 0,15 0,15 0,15
\(a.m_{H_2SO_4}=n.M=0,15.98=14,7\left(g\right)\)
\(b.m_{ddH_2SO_4}=D.V=50.1,2=60\left(g\right)\)
\(\rightarrow C\%_{ddH_2SO_4}=\frac{m_{ct}}{m_{dd}}.100\%=\frac{14,7}{60}.100\%=24,5\left(\%\right)\)
\(c.m_{ddspu}=m_{MgO}+m_{ddH_2SO_4}=6+60=66\left(g\right)\)
\(C\%_{ddM}=\frac{120,0,15}{66}.100\%=27,27\left(\%\right)\)
MgO + H2SO4 → MgSO4 + H2O
\(n_{MgO}=\frac{6}{40}=0,15\left(mol\right)\)
a) Theo PT: \(n_{H_2SO_4}=n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15\times98=14,7\left(g\right)\)
b) \(m_{ddH_2SO_4}=50\times1,2=60\left(g\right)\)
\(C\%_{H_2SO_4}=\frac{14,7}{60}\times100\%=24,5\%\)
c) \(m_{dd}saupư=6+60=66\left(g\right)\)
Theo Pt: \(n_{MgSO_4}=n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgSO_4}=0,15\times120=18\left(g\right)\)
\(\Rightarrow C\%_{MgSO_4}=\frac{18}{66}\times100\%=27,27\%\)
MgO + H2SO4 --> MgSO4 + H2O
a) nMgO = \(\frac{6}{40}\) = 0,15 mol
Theo PTHH, ta có: n\(H_2SO_4\) = 0,15 mol
m\(H_2SO_4\) = 0,15 . 98 = 14,7 g
b) mdd\(H_2SO_4\) = d.V = 1,2. 50 = 60 g
C%dd\(H_2SO_4\) = \(\frac{14,7}{60}.100\%=\) 24,5%
c) mdd\(MgSO_4\) (sau pư) = 6 + 60 = 66g
Theo PTHH , ta có: n\(MgSO_4\) = 0,15 mol
m\(MgSO_4\) = 0,15.120 = 18 g
C% = \(\frac{18.100\%}{66}\) = 27,27%
