1) Answer is: there is 0,554 moles in 20,2 grams of HCl.
m(HCl) = 20,2 g.
M(HCl) = 1,01 g/mol + 35,45 g/mol.
M(HCl) = 36,46 g/mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 20,2 g ÷ 36,46 g/mol.
n(HCl) = 0,554 mol.
n - amount of substance.
2) Answer is: there is 72,92 grams in 2 moles of HCl.
n(HCl) = 2 mol.
M(HCl) = 1,01 g/mol + 35,45 g/mol.
M(HCl) = 36,46 g/mol.
m(HCl) = n(HCl) · M(HCl).
n(HCl) = 2 mol · 36,46 g/mol.
n(HCl) = 72,92 g.
M - molar mass.
3) Answer is: there is 1,385·10²⁵ atoms ins 23 moles of sodium.
n(Na) = 23 mol.
N(Na) = n(Na) · Na.
N(Na) = 23 mol · 6,022·10²³ 1/mol.
N(Na) = 1,385·10²⁵.
Na - Avogadro number.
N - number of particles.
4) Answer is: there is 1332 moles in 8,022·10²⁶ molecules of NaCl.
N(NaCl) = 8,022·10²⁶.
N(NaCl) = n(Na) · Na.
n(NaCl) = N(NaCl) ÷ Na.
n(NaCl) = 8,022·10²⁶ ÷ 6,022·10²³ 1/mol.
n(NaCl) = 1,332·10³ mol = 1332 mol.
Na - Avogadro number.
N - number of particles.
5) Answer is: there is 0,223 moles in 5 liters of hydrogen gas.
V(H₂) = 5 L.
n(H₂) = ?.
Make proportion: 5 L : n(H₂) = 22,4 L : 1 mol.
22,4 · n(H₂) = 5 L · 1 mol.
n(H₂) = 0,223 mol.
V - volume of hydrogen gas.
6) Answer is: there is 112 liters in 5 moles of hydrogen gas.
n(H₂) = 5 mol.
V(H₂) = ?.
Make proportion: 5 mol : V(H₂) = 1 mol : 22,4 L
22,4 L · 5 mol = V(H₂) · 1 mol.
V(H₂) = 112 L.
V - volume of hydrogen gas.
n - amount of substance.
7) Answer is: there is 584,81 grams of HCl.
Chemical reaction: H₂ + Cl₂ → 2HCl.
m(H₂) = 16,2 g.
n(H₂) = m(H₂) ÷ M(H₂).
n(H₂) = 16,2 g ÷ 2,02 g/mol.
n(H₂) = 8,02 mol.
From chemical reaction: n(H₂) : n(HCl) = 1 : 2.
n(HCl) = 16,04 mol.
m(HCl) = 16,04 mol · 36,46 g/mol.
m(HCl) = 584,81 g.
8) Answer is: there is 14,51 liters in 46 g of Cl₂ gas.
m(Cl₂) = 46 g.
n(Cl₂) = m(Cl₂) ÷ M(Cl₂).
n(Cl₂) = 46 g ÷ 70,9 g/mol.
n(Cl₂) = 0,648 mol.
Make proportion: 0,648 mol : V(Cl₂) = 1 mol : 22,4 L.
V(Cl₂) = 0,648 mol · 22,4 L ÷ 1 mol.
V(Cl₂) = 14,51 L.