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Ronch [10]
3 years ago
6

If 638.44g CuSO4 reacts with 240.0g NaOH, which is the limiting reactant?

Chemistry
2 answers:
Cloud [144]3 years ago
4 0
CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄

M(CuSO₄)=159.61 g/mol
m(CuSO₄)=638.44 g
n(CuSO₄)=m(CuSO₄)/M(CuSO₄)
n(CuSO₄)=638.44/159.61= 4.00 mol

M(NaOH)=40.00 g/mol
m(NaOH)=240.0 g
n(NaOH)=m(NaOH)/M(NaOH)
n(NaOH)=240.0/40.00= 6.00 mol

on the reaction equation
CuSO₄ : NaOH = 1 : 2

in practice
CuSO₄ : NaOH = 4 : 6 = 1 : 1.5 ⇒ NaOH is the limiting reactant
Blizzard [7]3 years ago
3 0

Answer:

NaOH is the limiting reactant

Explanation:

Considering the chemiacl reaction :

CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄

Molar Mass CuSO₄ = 159 g/mol

Molar mass NaOH = 40 g/mol

Due to stoichiometry :

159 g CuSO₄ = 80 g NaOH

Then: 638.44 g CuSO₄ should react with 321.23 g NaOH, but we have 240.o g of NaOH, so this last is the limiting reactant

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a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
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Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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