Question

Calculează numărul de moli din: a) 64 g NaOH; b) 64g CuSO4; c) 80 g CH4
Calculează masa a: a) 3,5 moli MgO; b) 5 moli CaO; c) 8 moli NaNO3.
Cîti moli și câte molecule se gasesc în 73g HCl?

Answer 1

nr.moli = n = masa(m)/masa.moleculara(M)

a) M.NaOH = 40 g/mol

=> n = 64/40 = 1,6 moli NaOH

b) M.CuSO4 = 160 g/mol

=> n = 64/160 = 0,4 moli CuSO4

c) M.CH4 = 16 g/mol

=> n = 80/16 = 5 moli CH4

a) M.MgO = 40 g/mol

=> m = nxM = 40x3,5 = 140 g MgO

b) M.CaO = 56 g/mol

=> m = nxM = 56x5 = 280 g CaO

c) M.NaNO3 = 85 g/mol

=> m = nxM = 85x8 = 680 g NaNO3

M.HCl = 36,5 g/mol

1 mol HCl ........... 36,5 g ................. 6,022x10la23 molecule

x moli ................... 73 g ..................... y

= 2 moli HCl                                    = 1,2044x10la24 molecule

Answer 2

Răspuns:

Explicație:

a) ν NaOH= 23+16+1=40g/mol

1 mol NaOH... 40g

x mol NaOH.....64g

x= (64g*1mol)/40g= 1,6 mol

νCuSO4= 64+32+4*16=160g/mol

1 mol CuSO4.....160g

y mol CuSO4.....64g

y=(64g*1mol)/160g= 0,4 mol

νCH4= 12+4*1=16g/mol

1 mol CH4....16g

z mol CH4.....80g

z=(80g*1mol)/16g= 5 mol

b) νMgO= 24+16=40g/mol

1 mol MgO....40g

3,5 mol MgO....xg

x= (3,5mol*40g)/1mol= 140g

νCaO= 40+16=56g/mol

1 mol CaO....56g

5 mol CaO.....yg

y=(5mol*56g)/1mol=280g

νNaNO3=23+14+3*16=85g/mol

1 mol NaNO3....85g

8 mol NaNO3....zg

z=(8mol*85g)/1mol= 680g

c) νHCl= 1+35,5=36,5g/mol

1 mol HCl.....36,5g

x mol HCl.....73g

x=(73g*1mol)/36,5g=2 mol

1 mol HCl.......6,022*10 la puterea 23 molecule HCl

2 mol HCl,,,,,,y molecule HCl

y=(2mol*6,022)/1mol= 12,044* 10 la puterea 23 molecule HCl