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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: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{8}{64}=0,125\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\n_{H_2}=\dfrac{m}{M}=\dfrac{0,1}{2}=0,05\left(mol\right)\end{matrix}\right.\)
`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}+n_{H_2}=0,125+0,1+0,05=0,275\left(mol\right)\)
`=>` \(V_{hh\left(\text{đ}ktc\right)}=n_{hh}.22,4=0,275.22,4=6,16\left(l\right)\)
\(a,m_{CaSO_4}=136.0,25=34\left(g\right)\\ b,n_{Cu_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ m_{Cu_2O}=0,5.144=72\left(g\right)\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ m_{NH_3}=17.0,3=5,1\left(g\right)\\ d,m_{C_4H_{10}}=0,17.58=9,86\left(g\right)\\ e,n_{Cu\left(OH\right)_2}=\dfrac{4,5.10^{25}}{6.10^{23}}=75\left(mol\right)\\ m_{Cu\left(OH\right)_2}=98.75=7350\left(g\right)\\ g,m_{MgO}=0,48.40=19,2\left(g\right)\\ h,n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{CO_2}=44.0,15=6,6\left(g\right)\\ i,m_{Al\left(OH\right)_3}=78.0,25=19,5\left(g\right)\\\)
Các câu còn lại em làm tương tự nha!
\(a.m_O=1.16=16\left(g\right)\\ m_{O_2}=1.32=32\left(g\right)\\ b.m_{Fe}=1,5.56=84\left(g\right)\\ m_{Fe_2O_3}=1,5.160=240\left(g\right)\\ c.m_N=0,25.14=3,5\left(g\right)\\ m_{NO_2}=2,5.46=115\left(g\right)\\ d.m_{C_6H_{12}O_6}=1.180=180\left(g\right)\)
a) \(m_S=n_S.M_S=0,5.32=16\left(g\right)\)
b) \(m_{N_2}=n_{N_2}.M_{N_2}=1,5.28=42\left(g\right)\)
c) \(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,25.102=25,5\left(g\right)\)
d) \(n_H=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ \Rightarrow m_H=n_H.M_H=0,5.1=0,5\left(g\right)\)
e) \(n_{O_2}=\dfrac{V_{O_2\left(đktc\right)}}{22,4}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ \Rightarrow m_{O_2}=n_{O_2}.M_{O_2}=1,5.32=48\left(g\right)\)
f) \(n_{SO_3}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ \Rightarrow m_{SO_3}=n_{SO_3}.M_{SO_3}=1,5.80=120\left(g\right)\)