Question

рассчитайте хватит ли 8 г серы для реакции с 14 г цинка

Answer 1

Zn + S = ZnS
количество вещ.(S)=m/M= 8/32=0.25 моль
количество вещ.(Zn)= 0.25*1/1=0.25 моль
масса (Zn)= 0,25*65=16,25 гр
серы хватит.

Answer 2

Calculation of Sulfur and Zinc Reaction

To calculate whether 8g of sulfur is enough to react with 14g of zinc, we can use stoichiometry to determine the limiting reactant and the amount of product formed.

The balanced chemical equation for the reaction between sulfur and zinc is:

Zn + S → ZnS

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

Calculation Steps

1. Calculate the moles of sulfur and zinc. 2. Determine the limiting reactant. 3. Calculate the moles of product formed.

Step 1: Calculate the Moles of Sulfur and Zinc

The molar mass of sulfur (S) is approximately 32.07 g/mol, and the molar mass of zinc (Zn) is approximately 65.38 g/mol.

Moles of sulfur = Mass of sulfur / Molar mass of sulfur Moles of sulfur = 8g / 32.07 g/mol Moles of sulfur ≈ 0.249 moles

Moles of zinc = Mass of zinc / Molar mass of zinc Moles of zinc = 14g / 65.38 g/mol Moles of zinc ≈ 0.214 moles

Step 2: Determine the Limiting Reactant

To determine the limiting reactant, we compare the moles of sulfur and zinc. The limiting reactant is the one that is completely consumed in the reaction.

Since sulfur and zinc react in a 1:1 mole ratio, sulfur is the limiting reactant because there are fewer moles of sulfur compared to zinc.

Step 3: Calculate the Moles of Product Formed

The moles of product formed will be equal to the moles of the limiting reactant, which is sulfur.

Moles of product = Moles of limiting reactant Moles of product ≈ 0.249 moles

Conclusion

Based on the calculation, 8g of sulfur is enough to react with 14g of zinc. The limiting reactant is sulfur, and approximately 0.249 moles of zinc sulfide will be formed.

(#1)