Magnesium sulphide was obtained from 4-00 g of magnesium and 2-00 g of sulphur by the reaction Mg-S - MgS- The volume of oxygen at STP 1 atm- 273 K required for the complete oxidation of the excess reagent is-

Moles of Mg = 224 = 112 moles of S = 432 = 18 Finding the limiting reagent, nbsp;molesstoichiometric coefficient moles of Mg1 = 112 moles of S1 = 18 Hence ...

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Standard XI

Chemistry

Stoichiometric Calculations

Magnesium sul...

Question

Magnesium sulphide was obtained from $4.00 g$ of magnesium and $2.00 g$ of sulphur by the reaction $M g + S \to M g S$ . The volume of oxygen at STP ( $1 a t m, 273 K$ ) required for the complete oxidation of the excess reagent is:

0.93 L

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0.53 L

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0.63 L

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1.3 L

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Solution

The correct option is A 0.93 L
moles of $M g =$ $\frac{2}{24} = \frac{1}{12}$
moles of $S =$ $\frac{4}{32} = \frac{1}{8}$

Finding the limiting reagent,
$\frac{moles}{stoichiometric coefficient}$

$\frac{moles of Mg}{1} = \frac{1}{12}$
$\frac{moles of S}{1} = \frac{1}{8}$
Hence $M g$ is the limiting reagent.
Excess moles of $S$ left = $\frac{1}{8} - \frac{1}{12} = \frac{1}{24}$
oxidation of S,
$S + O_{2} \to S O_{2}$

By unitary method,
$\frac{moles of S}{1} = \frac{moles of O_{2}}{1}$
Moles of oxygen = $\frac{1 \times 1}{24 \times 1} = \frac{1}{24}$
Volume of oxygen at STP = $22.4 \times \frac{1}{24} = 0.93 L$

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Similar questions

Q. Magnesium sulphide was obtained from

4.00 g

of magnesium and

2.00 g

of sulphur by the reaction

M g + S \to M g S

. The volume of oxygen at STP (

1 a t m, 273 K

) required for the complete oxidation of the excess reagent is:

Q. How much magnesium sulphide can be obtained from

2.00

g of

M g

and

2.00

g of

S

by the following reaction.

M g + S \to M g S

Q. What volume of air at

1 a t m

and

273 K

containing 21% of oxygen by volume is required to completely burn sulphur

(S_{8})

present in

200 g

of sample, which contains

20 %

inert material which does not burn. Sulpur burns according to the reaction

\frac{1}{8} S_{8} (s) + O_{2} (g) \to S O_{2} (g)

Q. Students perform an experiment in which the mass of a sample of

M g

is vigorously heated so that it reacts with oxygen to form magnesium oxide. The purpose of the experiment is to verify the formula unit (

M g O

) of the compound produced in the reaction.

M g (s) + O_{2} (g) \to 2 M g O (s)

I

. The students find their initial sample of magnesium had a mass of

0.24 g

and the ending mass of the magnesium oxide was

0.48 g

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M g O

which matches perfectly with the ration predicted by the balanced equation.
BECAUSE

I I

. The magnesium and oxygen should react in a

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ratio.

Q. What volume of air at

1

atm and

273 K

that has 21% of oxygen by volume is required to completely burn sulphur

(S_{8})

present in

200 g

of a sample, which contains 20% inert material that does not burn? Sulphur burns according to the reaction:

\frac{1}{8} S_{8} (s) + O_{2} (g) \to S O_{2} (g)

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Magnesium sulphide was obtained from 4.00 g of magnesium and 2.00 g of sulphur by the reaction Mg+S→MgS. The volume of oxygen at STP (1 atm, 273 K) required for the complete oxidation of the excess reagent is:

Magnesium sulphide was obtained from $4.00 g$ of magnesium and $2.00 g$ of sulphur by the reaction $M g + S \to M g S$ . The volume of oxygen at STP ( $1 a t m, 273 K$ ) required for the complete oxidation of the excess reagent is: