3)a+b+c+d)

n_Zn=6,5:65=0,1(mol)

Zn+2HCl->ZnCl₂+H₂

0,1...0,2......0,1......0,1.....(mol)

Theo PTHH:n_HCl=0,2(mol)

Mà C_M(ddHCl)=2M

=>V=V_ddHCl=0,2:2=0,1(l)=100ml

\(V_{H_2}\)=0,1.22,4=2,24l

\(m_{ZnCl_2}\)=0,1.136=13,6(g)

2)a+b+c+d)

m_HCl=7,3%.50=3,65(g)

=>n_HCl=3,65:36,5=0,1(mol)

Zn+2HCl->ZnCl₂+H₂

0,05...0,1....0,05....0,05.......(mol)

Theo PTHH:m=m_Zn=0,05.65=3,25(g)

\(V_{H_2\left(đktc\right)}\)=0,05.22,4=1,12(l)

\(m_{ZnCl_2}\)=0,05.136=6,8(g)

1)n_CuO=4:80=0,05(mol)

n_HCl=2,92:36,5=0,08(mol)

a)+b)

CuO+2HCl->CuCl₂+H₂O

0,04.....0,08.....0,04............(mol)

Ta có:\(\dfrac{n_{CuO}}{1}>\dfrac{n_{HCl}}{2}\)=>CuO dư,HCl hết.Tính theo HCl.

Theo PTHH:

Sau pư,chất còn lại gồm:CuO dư,CuCl₂

Theo PTHH:m_CuO(dư)=(0,05-0,04).80=0,8(g)

\(m_{CuCl_2}\)=0,04.135=5,4(g)

Sửa đề:500g dd->500ml dd

\(n_{CO_2}\)=2,24:22,4=0,1(mol)

n_NaOH=2,5.0,5=1,25(mol)

Ta có:t=\(\dfrac{n_{NaOH}}{n_{CO_2}}\)=\(\dfrac{1,25}{0,1}\)=12,5>2=>Chỉ xảy ra phản ứng:

2NaOH+CO₂->Na₂CO₃+H₂O

...0,2..........0,1.........0,1...................(mol)

Ta có:\(\dfrac{n_{CO_2}}{1}< \dfrac{n_{NaOH}}{2}\)=>CO₂ hết.Tính theo CO₂

Theo PTHH:\(n_{Na_2SO_3}\)=0,1(mol)

n_NaOH(dư)=1,25-0,2=1,05(mol)

=>\(C_{M\left(Na_2CO_3\right)}\)=\(\dfrac{0,1}{0,5}\)=0,2M

=>C_M(NaOHdư)=1,05:0,5=2,1M

Số quá lớn đề có sai k vậy bạn

2)\(n_{CuSO_4.5H_2O}\)=25:250=0,1(mol)

=>\(n_{CuSO_4}\)=\(n_{CuSO_4.5H_2O}\)=0,1(mol)

=>\(m_{CuSO_4}\)=0,1.160=16(g)

1)Ở 20^oC

190g H₂O hòa tan tối đa 80g A tạo thành dd bão hòa

=>100g............................S g..................

Ta có:S_20^oC=\(\dfrac{100}{190}\).80=42,1(g)

a)CTHH:Fe(NO₃)₂

PTK=56+(14+16.3).2=180(đvC)

b)CTHH:Na₂SO₄

PTK=23.2+32+16.4=142(đvC)

c)CTHH:Ca₃(PO₄)₂

PTK=40.3+(31+16.4).2=310(đvC)

d)CTHH:CuCl₂

PTK=64+35,5.2=135(đvC)