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liubo4ka [24]
2 years ago
6

350.0-mL of 0.50 M hydrogen sulfate solution is reacted with 15.0 grams of sodium hydroxide. What volume of water will be produc

ed?​
Chemistry
1 answer:
Evgesh-ka [11]2 years ago
6 0

The volume of water that will be produced from the reaction will be 6.3 mL

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

H_2SO_4 + 2NaOH --- > Na_2SO_4 + 2H_2O

The mole ratio of hydrogen sulfate to sodium hydroxide is 1:2.

Mole of hydrogen sulfate = 0.50 x 350/1000 = 0.175 moles

Mole of 15 grams sodium hydroxide = 15/40 = 0.375 moles

Thus, hydrogen sulfide is the limiting reagent.

Mole ratio of hydrogen sulfide to water = 1:2.

Equivalent mole of water = 0.175 x 2 = 0.35 moles

Mass of 0.35 moles of water = 0.35 x 18 = 6.3 grams.

1 gram of water = 1 ml.

Thus, 6.3 grams of water will be equivalent to 6.3 mL

More on stoichiometric calculation can be found here: brainly.com/question/27287858

#SPJ1

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Explanation :

As we know that:

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At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between  pressure and number of moles of gas will be:

\frac{P_1}{P_2}=\frac{n_1}{n_2}

where,

P_1 = initial pressure of gas = 24.5 atm

P_2 = final pressure of gas = 5.30 atm

n_1 = initial number of moles of gas = 1.40 moles

n_2 = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}

n_2=0.301mol

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8 0
2 years ago
What are eight elements named after towns
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Convert this molecular formula into a structure that is consistent with the usual bonding patterns. C2h6o if more than one struc
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The given formula contains two carbon along with six hydrogen and an oxygen.

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8 0
3 years ago
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in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

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Difference between the freezing T° of pure solvent and freezing T° of solution

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So, if we replace in the formula

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These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

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