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2 years ago

What volume of a 0.452 m naoh solution is needed to neutralize 85.0 ml of a 0.176 m solution of h2so4?

Chemistry

2 answers:

ArbitrLikvidat [17]2 years ago

6 0

H2SO4 + 2 NaOH ----> Na2SO4 + 2 H2O.

0.085 L * 0.176 mol/L = 0.01496 mol H2SO4

is neutralised by 0.01496 mol * 2

= 0.02992 mol NaOH.

1000 mL of 0.492 M NaOH

contains 0.492 moles NaPH.

0.02992 / 0.452 * 1000 mL

= 66.19 = 66 mL

Akimi4 [234]2 years ago

6 0

Answer : The volume of $NaOH$ needed is, 45.4 mL

Explanation :

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.176M\\V_1=85.0mL\\n_2=1\\M_2=0.452M\\V_2=?$

Putting values in above equation, we get:

$2\times 0.176M\times 58.0mL=1\times 0.450M\times V_2\\\\V_2=45.4m:$

Hence, the volume of $NaOH$ needed is, 45.4 mL

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To calculate the mass of calcium carbonate for given moles, we use the equation:

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We are given:

One container contains eighty 1 g of tablets, this means that in total 80 g of tablets are there.

Every container has 40 % calcium carbonate.

Mass of calcium carbonate in 1 container = 40 % of 80 g = $\frac{40}{100}\times 80=32g$

Calculating the containers for amount of calcium carbonate that neutralized HCl by using unitary method:

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So, 10 g of calcium carbonate will be present in = $\frac{1}{32}\times 10=0.3125$ container

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