1) ai 4.8184 × [tex] 10^{24} [/tex] molecule daca raportam nr acesta la Nr lui Avogadro aflam nr de moli reprezentati de acel nr de molecule
anume: [tex] \frac{4.8184 * 10 ^{24}}{6.0223 * 10^{23} } = 0.8 mol H _{2} [/tex] => 0.8 mol * 2g/mol = 1.6 mol H2
8 mol [tex] N_{2} [/tex] = 8 x 28 g/mol = 224 g
2) 4 NH3 - amoniac, N(III), H(I)
3 CH4-metan, C(IV),H(I)
12 HCl -acid clorhidric, Cl (I), H(I)
6 H2O-apa- O(II), H(I)
6 H2S- acid sulfhidric, hidrogen sulfurat,S(II), H(I)
12 HF- acid fluorhidric F(I), H(I)
4 PH3- fosfina, P(III),H(I)
3 SiH4 - silan, tetrahidrura de siliciu, Si(IV), H(I)
trebuie sa ai 12 atomi de hidrogen (este nr cel mai mic pt a avea acelasi nr de H pt fiecare molecula)
3)X- Calciu (M=40)
Y- Clor atomic (M=35.5)
[tex] XY_{2} [/tex] = CaCl2
[tex]M_{CaCl_{2} [/tex]= 1 x Ca + 2x Cl = 1x 40 + 2x 35.5 = 40 + 71 = 111 g/mol
4 )
10 g H2O => n= 10 g : 18 g/mol = 0.55 mol
20 g NH3 => n= 20g : 17 g/mol = 1.18mol
10 g HCl = > n=10 g : 36.5 g/mol = 0.28 mol
n = [tex] \frac{m(g)}{M(g/mol)} [/tex]
5)
[tex] O_{2} +2H_2 =\ \textgreater \ 2 H_2O[/tex]
2 mol H2.... 2 mol H2O
X mol H2 ...0.55 mol H2O
X= 0.55 mol H2O x 2 / 2 = 0.55 mol H2
0.55 × 6.0223 ×[tex] 10^{23} [/tex] = .... atomi/mol
Identic se face la toate celelalte. Evident inlocuind reactantii pt NH3, HCl.
N2 + 3 H2 = > 2 HN3
H2 + Cl2 => 2 HCl
[tex]N_A [/tex] = 6.0223 × [tex]10^{23} [/tex] atomi/mol (Nr lui Avogadro)
Daca imparti nr de moli pe care ii ai tu la Nr lui Avogadro o sa iti dea nr de atomi din cantitatea ta de substanta
