Question

1. Tính số mol

a, 11,2 (g) Fe

b, 10 (g) \(CaCO_3\)

c, 5,6 (l) \(O_2\) (đktc)

d, 2 (l) NaOH 1M

e, 100ml HCL 0,1M

f, 100 (g) dung dịch NaOH 20%

( Fe = 56, CA = 40, C = 12, O = 16, Na = 23, H = 1 )

Answer 1

a. \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

b. \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

c. \(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

d. \(n_{NaOH}=1.2=2\left(mol\right)\)

e. \(n_{HCl}=0,1.100:1000=0,01\left(mol\right)\)

f. \(n_{NaOH}=\dfrac{20\%.100}{100\%}=20\left(mol\right)\)