$a,n_{Ca_3(PO_4)_2}=\dfrac{48}{310}\approx 0,155(mol)\\ b,n_{HCOOCH_3}=\dfrac{90}{60}=1,5(mol)\\ c,n_{Fe_2O_3}=\dfrac{40}{160}=0,25(mol)\\ d,n_{Al_2(SO_4)_3}=\dfrac{68,4}{342}=0,2(mol)\\ e,n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)\\ f,n_{Cl_2}=\dfrac{36.1}{(273+20).0,082}\approx 1,498(mol)\\ g,n_{CO_2}=\dfrac{56}{22,4}=2,5(mol)\\ h,n_{SO_2}=\dfrac{6.1}{(273+20).0,082}\approx 0,25(mol)\\ j,n_{K}=\dfrac{3.10^{23}}{6.10^{23}}=0,5(mol)\\ k,n_{CaSO_4}=\dfrac{12.10^{23}}{6.10^{23}}=2(mol)\\ l,n_{Na_2O}=\dfrac{6.10^{23}}{6.10^{23}}=1(mol)\\ m,n_{Zn}=\dfrac{18.10^{23}}{6.10^{23}}=3(mol)$

 

Đáp án:

Giải thích các bước giải:

a/ $n_{Ca_3(PO_4)_2}$ = $\frac{m}{M}$ = $\frac{48}{310}$ ≈ 0,15 mol

b/ $n_{HCOOCH_3}$ = $\frac{m}{M}$ = $\frac{90}{60}$ = 1,5 mol

c/ $n_{Fe_2O_3}$ = $\frac{m}{M}$ = $\frac{40}{160}$ = 0,25 mol

d/ $n_{Al_2(SO_4)_3}$ = $\frac{m}{M}$ = $\frac{68,4}{342}$ = 0,2 mol

e/ $n_{H_2}$ = $\frac{V}{22,4}$ = $\frac{13,44}{22,4}$ = 0,6 mol

f/ $n_{Cl_2}$ = $\frac{V}{24}$ = $\frac{36}{24}$ = 1,5 mol

g/ $n_{CO_2}$ = $\frac{V}{22,4}$ = $\frac{56}{22,4}$ = 2,5 mol

h/ $n_{SO_3}$ = $\frac{V}{24}$ = $\frac{6}{24}$ = 0,25 mol

i/ $n_{K}$ = $\frac{3.10^{23}}{6.10^{23}}$  = 0,5 mol

k/ $n_{CaSO_4}$ = $\frac{12.10^{23}}{6.10^{23}}$ = 2 mol

l/ $n_{Na_2}$O = $\frac{6.10^{23}}{6.10^{23}}$ = 1 mol

m/ $n_{Zn}$ = $\frac{18.10^{23}}{6.10^{23}}$ = 3 mol

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