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\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)
a) \(Pt:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) \(n_{Fe}=\dfrac{0,56}{56}=0,01mol\)
Theo pt: \(n_{FeSO_4}=n_{Fe}=0,01mol\)
\(\Rightarrow m_{FeSO_4}=0,01.152=1,52g\)
Theo pt: \(n_{H_2}=n_{Fe}=0,01mol\)
\(\Rightarrow V_{H_2}=0,01.22,4=0,224lít\)
c) \(Theopt:nH_2SO_4=n_{Fe}=0,01mol\)
\(\Rightarrow m_{H_2SO_4}=0,01.98=0,98g\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,98.100}{19,6}=5g\)
a) n N2=\(\frac{0.54.10^{23}}{6.10^{23}}=0,09.10^{23}\)
b) n CO2=\(\frac{4,4}{44}=0,1\left(mol\right)\)
V CO2=0,1.22,4=2,24(l)
c) n N2O5=\(\frac{8,961}{22,4}=0,4\left(mol\right)\)
m N2O5=0,4.108=43,2(g)
d) m Fe(OH)3=1,5.107=160,5(g)
\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ m_{H_2}=1,5.2=3\left(g\right)\)
PTHH : 2Al + H2SO4 -> Al2SO4 + H2
Theo ĐLBTKL
\(m_{Al}+m_{H_2SO_4}=m_{Al_2SO_4}+m_{H_2}\\ \Rightarrow m_{H_2SO_4}=\left(171+3\right)-2,7=171,3\left(g\right)\)
pthh: 2Al+3H\(_2\)SO\(_4\)→Al\(_2\)(SO4)\(_3\)+3H\(_2\)↑
nH\(_2=33,6:22,4=1,5\left(mol\right)\)
\(mH_2=1,5.2=3\left(g\right)\)
\(nAl_2\left(SO_4\right)=171:150=1,14\left(mol\right)\)
\(mAl_2\left(SO_4\right)_3=1,14.342=389,88\left(g\right)\)
BTKL : mAl + mH\(_2\)SO\(_4\) = m Al\(_2\)(SO4)\(_3\) + m H\(_2\)
2,7 + mH\(_2\)SO\(_4\) = 389,88 + 3
=> \(mH_2SO_4=\left(389,88+3\right)-2,7=390,18\left(g\right)\)
a)
n Fe2O3=8/(56✖ 2+16❌ 3)=0.05mol
b)nH2=6.72/22.4=0.3mol
c) nH2=9.1023/6.1023=1.5mol
VH2=1,5✖ 22.4=33.6l
d)nO2=3,2/32=0,1mol
➡ nN2=0,4mol
mN2=0,4✖ 28=11,2g
e)nFe2(SO4)3=8/400=0,02mol
f)nH2=(1,2✖ 10^23)/6✖ 10^23=0,2mol
nN2=2,8/28=0,1mol
VN2=0,1✖ 22,4=2,24l
VO2=1,5✖ 22,4=33,6l
VH2=0,1 ✖ 22,4=2,24l
VX=2,24+2,24+33,6=38.08l
mO2=1,5❌ 32=48g
mN2=0,1✖ 28=2,8g
mH2=0,1✖ 2=0,2g
mX=2,8+0,2+48=51g
a)
- \(V_{CO}=n.24=0,2.24=4,8\left(l\right)\)
- \(n_{SO_3}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\)
`=>` \(V_{SO_3}=n.24=0,1.24=2,4\left(l\right)\)
- \(n_{N_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
`=>` \(V_{N_2}=n.24=0,5.24=12\left(l\right)\)
b)
- \(m_{Fe_2O_3}=n.M=0,25.160=40\left(g\right)\)
- \(m_{Al_2O_3}=n.M=0,15.102=15,3\left(g\right)\)
- \(n_{O_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
`=>` \(m_{O_2}=n.M=0,15.32=4,8\left(g\right)\)
c)
Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{8}{64}=0,125\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\n_{H_2}=\dfrac{m}{M}=\dfrac{0,1}{2}=0,05\left(mol\right)\end{matrix}\right.\)
`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}+n_{H_2}=0,125+0,1+0,05=0,275\left(mol\right)\)
`=>` \(V_{hh\left(\text{đ}ktc\right)}=n_{hh}.22,4=0,275.22,4=6,16\left(l\right)\)
a) \(n_{H_{2}SO_{4}}\)= \(\dfrac{0,3.10^{23}}{6.10^{23}}\)=0,05mol.
\(m_{H_{2}SO_{4}}\)= 0,05.98=4,9g
b)\(V_{CO_{2}}\)= 0,25.22,4=5,6l