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1)

a) mFe2O3 = 0.15*160 = 24 g

b) mMgO = 0.75*40=30 g

2)

a) nCuO = 2/80 = 0.025 mol

b) nNaOH = 10/40 = 0.25 mol

\(m_{Fe}=0,5.56=28\left(g\right)\)

\(m_{H_2SO_4}=0,75.98=73,5\left(g\right)\)

\(m_{CuSO_4}=0,6.160=96\left(g\right)\)

a. \(n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

PTHH : Fe₂O₃ + 3H₂ -> 2Fe + 3H₂O

           0,5          1,5         1

\(m_{Fe_2O_3}=0,5.160=80\left(g\right)\)

b. \(m_{Fe}=1.56=56\left(g\right)\)

nH2 = 9.10^23/6.10^23 = 1,5 (mol)

PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

Mol: 0,5 <--- 1,5 ---> 1

mFe2O3 = 0,5 . 160 = 80 (g)

mFe = 1 . 56 = 56 (g)

Fe₂O₃+3H₂-to>2Fe+3H₂O

0,5-------1,5---------1 mol

n _H2=\(\dfrac{9,10^{23}}{6.10^{23}}\)=1,5 mol

=>m _Fe3O4=0,5.232=116g

=>m _Fe=1.56=56g

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

V_X=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;

m_Al=27.0,9=24,3(g)

V_Al=24,3:2,7=9(cm³)

b;

m_Cl2=71.1,25=88,75(g)

V_Cl2=22,4.1,25=28(lít)

c;

m_NH3=0,86.17=14,62(g)

V_NH3=0,86.22,4=19,264(lít)

\(a.m_{H_2SO_4}=0,75.98=73,5g\)

\(b.m_{Cu}=0,6.64=38,4g\\ c.m_{\left(NH_4\right)_2SO_4}=0,8.132105,6g\)

\(n_{FeSO_4}=\dfrac{22,8}{56+32+16.4}=0,15(mol)\)

Chọn B

\(n_{FeSO_4}=\dfrac{22,8}{152}=0,15\left(mol\right)\)

=> B

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)