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3 years ago

What is the volume of 15.2 grams of SO2 at STP?

Chemistry

1 answer:

kiruha [24]3 years ago

5 0

5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.

Explanation:

Data given:

mass of sulphur dioxide = 15.2 grams

conditions is at STP whech means volume = 22.4 litres

atomic mass of sulphur dioxide = 64.06 grams/mole

Number of moles is calculated as:

number of moles = $\frac{mas}{atomic mass of 1 mole}$

Putting the values in the equation:

number of moles = $\frac{15.2}{64.06}$

= 0.23 moles

Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:

When 1 mole of sulphur dioxide occupies 22.4 litres at STP

Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23

= 5.152 litres is the volume.

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Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Write a balanced chemical equation for t

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Answer:

The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2

The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.

The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.

Explanation:

First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.

Mg + HCL ⇒ MgCl2 + H2

For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.

The atomic mass of Mg = 24.31 grs/mol

24.31 grs------------ 1 mol Mg

0.510 grs------------ x=0.02 mol Mg.

If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with

0.04 mol HCl. So, the molarity of the solution is 0.04 M HCl.

Then to calculate the pH we use the formula pH = - log [H+]

⇒ pH = -log [0.04]⇒ pH=1.4.

Finally, from the balanced equation and the findings described, and considering that at 25°C and 1.00 atm 1 mol of gas has volume of 24.1 L.

1 mol H2----------- 24.1 L

0.02 mol H2----- x= 0.482L.

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