Question

Tính số mol của chất sau:

500 ml dung dịch NaOH 20% (d = 1/2 g/ml)

200 gam dd H₂SO₄ 5M ( d = 1,29 g/ml)

Answer 1

a)

m\(_{NaOH}=\)\(500.\frac{1}{2}=250\left(g\right)\)

\(m_{NaOH\left(20\%\right)}=\frac{250.20}{100}=50\left(g\right)\)

\(n_{NaOH}=\frac{50}{40}=1,25\left(mol\right)\)

b)

\(V_{H2SO4}=\frac{200}{1,29}=155\left(ml\right)=0,155\left(l\right)\)

n\(_{H2SO4}=0,155.5=0,775\left(mol\right)\)

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Answer 2

\(a.\\ m_{NaOH}=\frac{500.1}{2}=250\left(g\right)\\ \rightarrow n_{NaOH}=\frac{250.20}{100.40}=1,25\left(mol\right)\\ b.\\ V_{dd}=\frac{200}{1,29}=155\left(ml\right)=0,155\left(l\right)\\ n_{axit}=0,155.5=0,775\left(mol\right)\)

Answer 3

Mdd NaOH= 500. ½ = 250(g)

=> M NaOH = 250. 20% = 50(g)

=> n NaOH= 50/40= 1,25 mol

V dd H2SO4 = 200: 1,29 = 20000/129(ml)

= 20/129(l)

=> N H2SO4 = 20/129 .5 \(\approx\)0, 775(mol)