Question

Tính thể tích của":

a)0,2mol khí H₂(đktc)

b)0,1mol khí CO₂(đktc)

c) 7,1g khí Clo(đktc)

d) 6,4 g khí Oxi(đktc)

e) 4,4g khí CO₂(đktc)

Answer 1

Tính thể tích của":

a)0,2mol khí H2(đktc)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b)0,1mol khí CO2(đktc)

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

c) 7,1g khí Clo(đktc)

\(V_{Cl_2}=\frac{7,1}{71}.22,4=2,24\left(l\right)\)

d) 6,4 g khí Oxi(đktc)

\(V_{O_2}=\frac{64}{32}.22,4=4,48\left(l\right)\)

e) 4,4g khí CO2(đktc)

\(V_{CO_2}=\frac{4,4}{44}.22,4=2,24\left(l\right)\)

Answer 2

a/

\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)(đktc)

b/

\(V_{CO_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)(đktc)

c/\(n_{Cl}=\frac{m}{M}\frac{7,1}{35,5}=0,2\left(mol\right)\)

➞\(V_{Cl}=n.22,4=0,2.22,4=4,48\left(l\right)\)(đktc)

d/

\(n_{O_2}=\frac{m}{M}=\frac{6,4}{32}=0,2\left(mol\right)\)

➞\(V_{O_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)(đktc)

e/\(n_{CO_2}=\frac{4,4}{44}=0,1\left(mol\right)\)

➞\(V_{CO_2}=n.22,4=0,1.22,4=2,24\left(l\right)\left(đktc\right)\)

Answer 3

a)

VH2 = 0.2*22.4 = 4.48 l

b) VCO2 = 0.1*22.4 = 2.24 l

c)

nCl2 = 7.1/71=0.1 mol

VCl2 = 0.1*22.4 = 2.24 l

d)

nO2 = 6.4/32 = 0.2 mol

VO2 = 0.2*22.4 = 4.48l

e)

nCO2 = 4.4/44= 0.1 mol

VCO2 = 0.1*22.4= 2.24 l