Question

Giup minh giai minh cau nay voi

Answer 1

1_)

PTHH :

\(Na2CO3+Ca\left(OH\right)2->2NaOH+CaCO3\)

0,25mol..........................................0,25mol

=> mNaOH( điều chế được) = 0,25.40 = 10(g)

2_)

Theo đề ta có : mNaCl = \(\dfrac{58,5.50}{100}=29,25\left(kg\right)\)

PTHH :

\(2NaCl+2H2O\xrightarrow[c\text{ó}-m\text{àng}-ng\text{ắn}]{\text{đ}p\text{dd}}2NaOH+Cl2\uparrow+H2\uparrow\)

117kg--------------------------------> 80kg ----------------> 2kg

29,25kg-----------------------------> x kg ------------------> y kg

=> \(\left\{{}\begin{matrix}x=mNaOH=\dfrac{29,25.80}{117}=20\left(kg\right)\\y=mH2=\dfrac{29,25.2}{117}=0,5\left(kg\right)=>nH2=\dfrac{0,5.1000}{2}=250\left(mol\right)\end{matrix}\right.\)

=> VH2(đktc) = 250.22,4 = 5600 ( l)

Answer 2

a)PTHHH : Ca(OH)₂ + Na₂CO₃ \(\rightarrow\) CaCO₃ + 2NaOH

b) Theo PTHH ta có \(n_{NaOH}=2n_{Na_2CO_3}=0,25\cdot2=0.5\left(mol\right)\)

Số mol NaOH thu được từ 0,5 mol Natri Cabonat là

\(m_{Na_2CO_3}=M_{Na_2CO_3}\cdot n_{Na_2CO_3}=106\cdot0,5=53\left(g\right)\)