Question

Сколько граммов сульфида магния образуется при взаимодействии 64г. серы S с магнием Mg?(с решением)

Answer 1

X (Грамм)           64 (Грамма)
Mg            + S =        MgS
m = 24 г/моль     m = 56 г/моль
1 (Моль)            1 (Моль)

$\frac{x}{2} = \frac{64}{56}$

2 (Грамма)

Ответ: 2 (Грамма)

Answer 2

Calculating the Amount of Magnesium Sulfide Formed

To calculate the amount of magnesium sulfide formed when 64g of sulfur (S) reacts with magnesium (Mg), we can use the balanced chemical equation for the reaction between sulfur and magnesium. The balanced chemical equation for the reaction is:

S + Mg → MgS

From the balanced chemical equation, we can see that 1 mole of sulfur reacts with 1 mole of magnesium to produce 1 mole of magnesium sulfide.

The molar mass of sulfur (S) is approximately 32.06 g/mol, and the molar mass of magnesium (Mg) is approximately 24.31 g/mol. Using this information, we can calculate the amount of magnesium sulfide formed.

Calculation:

1. Calculate the number of moles of sulfur: - Number of moles = Mass / Molar mass - Number of moles = 64g / 32.06 g/mol - Number of moles = 1.993 moles

2. Since the reaction is 1:1 between sulfur and magnesium, the number of moles of magnesium is also 1.993 moles.

3. Now, calculate the mass of magnesium sulfide formed: - Mass = Number of moles * Molar mass - Mass = 1.993 moles * (24.31 g/mol + 32.06 g/mol) - Mass = 1.993 moles * 56.37 g/mol - Mass ≈ 112.33 grams

Therefore, approximately 112.33 grams of magnesium sulfide are formed when 64 grams of sulfur reacts with magnesium.