a) \(m_{Na}=n.M=0,3.23=6,9\left(g\right)\)

 \(m_{O_2}=n_{O_2}.M_{O_2}=0,3.32=9,6\left(g\right)\)

b) \(m_{HNO_3}=n_{HNO_3}.M_{HNO_3}=1,2.63=75,6\left(g\right)\)

    \(m_{Cu}=n.M=0,5.64=32\left(g\right)\)

c) \(m_{KNO_3}=n.M=0,125=0,125.101=12,625\left(g\right)\)

    \(m_{KMnO_4}=n.M=0,125.158=19,75\left(g\right)\)

     \(m_{KClO_3}=n.M=0,125.122,5=15,3125\left(g\right)\)

Câu 9. Tính khối lượng của những lượng chất sau:
a) 0,3 mol nguyên tử Na;=>m Na=0,3.23=6,9g

0,3 mol phân tử O2=>m O2=0,3.32=9,6g

b) 1,2 mol phân tử HNO3; =>n HNO3=1,2.63=75,6g

0,5 mol phân tử Cu=>m Cu=0,5.64=32g

c) 0,125 mol của mỗi chất sau:

KNO₃, KMnO₄, KClO₃ 

m KNO3=0,125.101=12,625g

m KMnO4==0,125.158=19,75g

m KClO3=0,125.122,5=15,3125g

a.

\(m_{H_2SO_4}=0.5\cdot98=49\left(g\right)\)

\(m_{NaOH}=0.2\cdot40=8\left(g\right)\)

\(m_{Ag}=0.1\cdot108=10.8\left(g\right)\)

b.

\(n_{SO_2}=\dfrac{15\cdot10^{23}}{6\cdot10^{23}}=2.5\left(mol\right)\)

\(m_{SO_2}=2.5\cdot64=160\left(g\right)\)

a . mA= 0,5x32=16 g

b . n=36x10²³/6x10²³=6(mol)

V=6x22,4=134,4 g

4.

a) \(V_{SO_2}=0.5\cdot22.4=11.2\left(l\right)\)

b) \(V_{CH_4}=\dfrac{3.2}{16}\cdot22.4=4.48\left(l\right)\)

c) \(V_{N_2}=\dfrac{0.9\cdot10^{23}}{6\cdot10^{23}}\cdot22.4=3.36\left(l\right)\)

5.

a) \(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

b) \(m_{Cu\left(NO_3\right)_2}=0.3\cdot188=56.4\left(g\right)\)

c) \(m_{Na_2CO_3}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}\cdot106=21.2\left(g\right)\)

d) \(m_{CO_2}=\dfrac{8.96}{22.4}\cdot44=17.6\left(g\right)\)

e) \(m_K=0.5\cdot2\cdot39=39\left(g\right)\\ m_C=0.5\cdot12=6\left(g\right)\\ m_O=0.5\cdot3\cdot16=24\left(g\right)\)

a) mN = 0,5 .14 = 7g.

mCl = 0,1 .35.5 = 3.55g

mO = 3.16 = 48g.

b) mN2 = 0,5 .28 = 14g.

mCl2 = 0,1 .71 = 7,1g

mO2 = 3.32 =96g

c) mFe = 0,1 .56 =5,6g mCu = 2,15.64 = 137,6g

mH2SO4 = 0,8.98 = 78,4g.

mCuSO4 = 0,5 .160 = 80g

a) nCO2=[(9.10²³)/(6.10²³)]=1,5(mol)

=> mCO2=1,5.44=66(g)

V(CO2,đktc)=1,5.22,4=33,6(l)

b) nH2=4/2=2(mol)

N(H2)=2.6.10²³=12.10²³(phân tử)

V(H2,đktc)=2.22,4=44,8(l)

c) N(CO2)=0,5.6.10²³=3.10²³(phân tử)

V(CO2,đktc)=0,5.22,4=11,2(l)

mCO2=0,5.44=22(g)

d) nN2=2,24/22,4=0,1(mol)

mN2=0,1.28=2,8(g)

N(N2)=0,1.10²³.6=6.10²² (phân tử)

e) nCu=[(3,01.10²³)/(6,02.10²³)]=0,5(mol)

mCu=0,5.64=32(g)

Mà sao tính thể tích ta :3