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mNaOH=100.5%=5g

Gọi mddNaOH thêm =x(g)

=>mNaOH=0,2x

Tổng mNaOH=0,2x+5

Tổng mdd=x+100

=>0,2x+5=(x+100)0,1

=>0,1x=5=>x=50g

Câu 2 : 

$n_{HCl} = 0,2.1 + 0,3.1,5 = 0,65(mol)$
$V_{dd} = 0,2 + 0,3 = 0,5(mol)$
$C_{M_{HCl}} = \dfrac{0,65}{0,5} =1,3M$

Câu 3 : 

Gọi $m_{H_2O\ cần\ thêm} =a (gam)$

Sau khi thêm : 

$m_{NaOH} = 100.35\% = 35(gam)$
$m_{dd} = 100 + a(gam)$

Suy ra:  $\dfrac{35}{100 + a}.100\% = 20\%$

Suy ra: a = 75(gam)

Câu 4 : 

Gọi $V_{dd\ HCl\ 2M} =a (lít) ; V_{dd\ HCl\ 3M} = b(lít)$

Ta có : 

$a + b = 4$
$2a + 3b = 4.2,75$

Suy ra a = 1(lít) ; b = 3(lít)

1.m_NaCl=200.10%+300.20%=80(g)

\(C\%=\dfrac{80}{200+300}.100\%=16\%\)

Khối lượng NaOH có trong 400g dd NaOH 20%:

\(m_{NaOH\left(1\right)}=\dfrac{400.20\%}{100\%}=80g\)

Có:

\(C_{\%NaOH,sau}=\dfrac{80+\dfrac{m.10\%}{100\%}}{400+m}\cdot100\%=12,5\%\\ \Rightarrow m=1200g\)

Sửa đề: "thu được dd NaOH có nồng độ phần trăm là?"

\(m_{KOH\left(20\%\right)}=20\%.120=24\left(g\right)\\ m_{KOH\left(10\%\right)}=280.100\%=28\left(g\right)\\ \rightarrow m_{KOH\left(\text{sau khi pha}\right)}=28+24=52\left(g\right)\\ m_{ddKOH\left(\text{sau khi pha}\right)}=120+280=400\left(g\right)\\ \rightarrow C\%_{KOH\left(\text{sau khi pha}\right)}=\dfrac{52}{400}=13\%\)

a)m dd sau=100gam
mNaCl không đổi=80.15%=12 gam
C% dd NaCl sau=12/100.100%=12%
b)mdd sau=200+300=500 gam
Tổng mNaCl sau khi trộn=200.20%+300.5%=55 gam
C% dd NaCl sau=55/500.100%=11%
c) mdd sau=150 gam
mNaOH trg dd 10%=5 gam
mNaOH trong dd sau khi trộn=150.7,5%=11,25 gam
=>mNaOH trong dd a%=11,25-5=6,25 gam
=>C%=a%=6,25/100.100%=6,25% => a=6,25
 

.//

a)m dd sau=100gam
mNaCl không đổi=80.15%=12 gam
C% dd NaCl sau=12/100.100%=12%
b)mdd sau=200+300=500 gam
Tổng mNaCl sau khi trộn=200.20%+300.5%=55 gam
C% dd NaCl sau=55/500.100%=11%
c) mdd sau=150 gam
mNaOH trg dd 10%=5 gam
mNaOH trong dd sau khi trộn=150.7,5%=11,25 gam
=>mNaOH trong dd a%=11,25-5=6,25 gam
=>C%=a%=6,25/100.100%=6,25% => a=6,25

Ta có: \(m_{CuSO_4}=\dfrac{10\%.160}{100\%}=16\left(g\right)\)

=> \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

Ta lại có: \(m_{NaOH}=\dfrac{5\%.240}{100\%}=12\left(g\right)\)

=> \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)

a. PTHH: \(CuSO_4+2NaOH--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\)

Vậy NaOH dư, CuSO₄ hết.

Theo PT: \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\)

=> \(m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

b. Ta có: \(m_{dd_{Na_2SO_4}}=240+16-9,8=246,2\left(g\right)\)

Ta có: \(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

=> \(C_{\%_{Na_2SO_4}}=\dfrac{14,2}{246,2}.100\%=5,77\%\)

Tiếp: 

Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(m_{NaOH_{PỨ}}=0,2.40=8\left(g\right)\)

=> \(m_{NaOH_{dư}}=12-8=4\left(g\right)\)

=> \(C_{\%_{NaOH_{dư}}}=\dfrac{4}{240}.100\%=1,7\%\)

\(a,m_{CuSO_4}=160.10\%=16\left(g\right)\\ n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\\ b,m_{NaOH}=240.5\%=12\left(g\right)\\ n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\\ PTHH:2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\\ Vì:\dfrac{n_{NaOH\left(đề\right)}}{n_{NaOH\left(PTHH\right)}}=\dfrac{0,3}{2}>\dfrac{n_{CuSO_4\left(đề\right)}}{n_{CuSO_4\left(PTHH\right)}}=\dfrac{0,1}{1}\\ \Rightarrow CuSO_4hết,NaOHdư\)

Vậy:

 \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

\(b,m_{ddB}=m_{ddCuSO_4}+m_{ddNaOH}-m_{\downarrow}=160+240-9,8=390,2\left(g\right)\)

\(n_{NaOH\left(dư\right)}=0,3-0,1.2=0,1\left(mol\right)\)

\(C\%_{ddNa_2SO_4}=\dfrac{0,1.142}{390,2}.100\approx3,639\%\\ C\%_{ddNaOH\left(dư\right)}=\dfrac{0,1.40}{390,2}.100\approx1,025\%\)