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\(a,m_{Fe_2O_3}=0,15.(56.2+16.3)=0,15.160=24\\ b,m_{MgO}=0,75.(24+16)=0,75.40=30\)
\(m_{Cl_2}=1.71=71\left(g\right)\)
\(m_{CH_4}=1.16=16\left(g\right)\)
\(m_{CO_2}=1.44=44\left(g\right)\)
\(m_{K_2O}=1.94=94\left(g\right)\)
\(m_{Fe_2O_3}=1.160=160\left(g\right)\)
\(m_{CuSO_4}=1.160=160\left(g\right)\)
\(m_{NaOH}=1.40=40\left(g\right)\)
\(m_{Fe\left(NO_3\right)_2}=1.242=242\left(g\right)\)
\(m_{Fe\left(OH\right)_2}=1.90=90\left(g\right)\)
\(m_{KNO_3}=1.101=101\left(g\right)\)
\(m_{CaCO_3}=0,1.100=10\left(g\right)\)
\(m_{H_2O}=0,5.18=9\left(g\right)\)
\(m_{CuO}=0,15.80=12\left(g\right)\)
Bài 1 :
a) \(n_{NaOH}=\dfrac{m_{NaOH}}{M_{NaOH}}=\dfrac{10}{40}=0,25\left(mol\right)\)
b) \(m_{CuO}=n_{CuO}\cdot M_{CuO}=0,15\cdot80=12\left(g\right)\)
1.
\(a,n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ n_{CuO}=\dfrac{8}{80}=0,1(mol)\\ n_{Fe_2O_3}=\dfrac{16}{160}=0,1(mol)\\ b,V_{CO_2}=0,25.22,4=5,6(l)\\ V_{H_2}=0,175.22,4=3,92(l)\\ V_{N_2}=1,5.22,4=33,6(l)\)
2.
\(a,n_{Al}=0,5.2=1(mol);n_{O}=0,5.3=1,5(mol)\\ \Rightarrow m_{Al}=1.27=27(g);m_{O}=1,5.16=24(g)\\ b,n_{CO_2}=\dfrac{2,2}{44}=0,05(mol)\\ \Rightarrow m_C=0,05.12=0,6(g);m_O=0,05.2.16=1,6(g)\)
Bài 3:
1. mMg= 0,5.24=12(g)
mZn= 0,5. 65= 32,5(g)
2. mN=0,3.14=4,2(g)
mO2=0,3.32=9,6(g)
3. nNH3= 2. 17=34(g)
mO2=32.2= 64(g/mol)
4. mMgO= 40. 0,4=16(g)
mAl2O3= 102. 0,4= 40,8(g)
5. mCaCO3=2,5.100=250(g)
mCuSO4= 2,5.160=400(g)
\(a.n_{CO_2}=\dfrac{18.10^{23}}{6.10^{23}}=3\left(mol\right)\\ \Rightarrow m_{CO_2}=44.3=132\left(g\right)\\ b.n_{H_2O}=\dfrac{39,6}{18}=2,2\left(mol\right)\\ c.n_{Fe}=\dfrac{12.10^{23}}{6.10^{23}}=2\left(mol\right)\)
\(n_{Fe_2O_3}=\dfrac{4}{160}=0,025\left(mol\right)\)
PTHH :
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
0,025 0,15 0,05 0,075
\(a,m_{FeCl_3}=0,05.162,5=8,125\left(g\right)\)
\(b,C_{M\left(HCl\right)}=\dfrac{0,15}{0,15}=1M\)
Câu 1:
\(m_{H_2S}=0,75.34=25,5(g)\\ m_{CaSO_4}=0,025.136=3,4(g)\\ m_{Fe_2O_3}=0,05.160=8(g)\)
Câu 2:
\(V_{N_2}=2,5.22,4=56(l)\\ V_{H_2}=0,03.22,4=0,672(l)\\ V_{O_2}=0,45.22,4=10,08(l)\\ V_{hh}=22,4.(0,2+0,25)=22,4.0,45=10,08(l)\)
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