Question

tính KL chất tan của:a)0,2mol NaCO3 b)200 ml dd NaOH1,5Mc)300g dd HCL 14,6%

 

Answer 1

a)

\(m_{Na_2CO_3}=n.M=0,2.106=21,2\left(g\right)\)

b)

Đổi:\(200ml=0,2l\)

\(n_{NaOH}=0,2.1,5=0,3\left(mol\right)\)

\(m_{NaOH}=0,3.40=12\left(g\right)\)

c)

\(m_{HCl}=\dfrac{300.14,6}{100}=43,8\left(g\right)\)

Answer 2

a, m_Na2CO3 = 0,2.106 = 21,2 (g)

b, n_NaOH = 0,2.1,5 = 0,3 (mol)

⇒ m_NaOH = 0,3.40 = 12 (g)

c, m_HCl = 300.14,6% = 43,8 (g)

Answer 3

\(a)m_{Na_2CO_3}=0,3.106=31,8g\\ b)n_{NaOH}=0,2.1,5=0,3mol\\ m_{NaOH}=0,3.40=12g\\ c)m_{HCl}=\dfrac{300.14,6\%}{100\%}=43,8g\)