Question

2Al + 6HCl = 2AlCl3 + 3H2 m(Al)=2,7 гНайти:m(HCl)V(H2)

Answer 1

2Al + 6HCl = 2AlCl3 + 3H2
n=m/M(Ar)  Ar(Al)=27  M=27г/моль
n(Al)=2,7/27=0,1 моль
n(Al) :  n(HCl)=2 : 6, след., в 3 раза больше n(HCl)=0,1*3=0,3моль, по условию реакции
m=n*M(Mr)    Mr(HCl)=36,5    М=36,5г/моль
m(HCl)= 0,3*36,5=10,95г
n(Al) : n(H2)=2 : 3, след., в 1,5 раза больше n(H2)= 0,1*1,5=0,15 моль, по условию реакции
V=n*Vm    V(H2)= 0,15*22,4=3,36л
Ответ: 10,95г    3,36л

Answer 2

To find the values of m(HCl) and V(H2) in the given chemical equation:

2Al + 6HCl = 2AlCl3 + 3H2

m(Al) = 2.7 g (given)

From the balanced equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 3 moles of H2. Therefore, we need to determine the moles of HCl to find m(HCl) and then use the stoichiometry to calculate V(H2).

Molar mass of Al (Aluminum) = 26.98 g/mol Molar mass of HCl (Hydrochloric acid) = 36.46 g/mol

1. Calculate the moles of Al: moles of Al = m(Al) / Molar mass of Al moles of Al = 2.7 g / 26.98 g/mol moles of Al ≈ 0.1 mol

2. Use the mole ratio from the balanced equation to determine the moles of HCl: moles of HCl = 6 * moles of Al moles of HCl = 6 * 0.1 mol moles of HCl = 0.6 mol

3. Calculate the mass of HCl: m(HCl) = moles of HCl * Molar mass of HCl m(HCl) = 0.6 mol * 36.46 g/mol m(HCl) ≈ 21.88 g

4. Use the mole ratio from the balanced equation to determine the moles of H2: moles of H2 = 3 * moles of Al moles of H2 = 3 * 0.1 mol moles of H2 = 0.3 mol

5. Calculate the volume of H2 at standard temperature and pressure (STP): 1 mole of any gas at STP occupies 22.4 liters.

V(H2) = moles of H2 * 22.4 L/mol V(H2) = 0.3 mol * 22.4 L/mol V(H2) = 6.72 L

Therefore, the values are: m(HCl) ≈ 21.88 g V(H2) = 6.72 L