? ZnS+?O2 !? ZnO+? SO2 ,
what is the maximum amount of ZnO
(81.4084 g/mol) which could be formed from
12.85 g of ZnS (97.474 g/mol) and 15.91 g of
O2 (31.9988 g/mol)?
Answer in units of g
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To find the maximum amount of ZnO that can be formed, we need to determine the limiting reactant in the given chemical equation. The limiting reactant is the reactant that will be completely consumed and determines the maximum amount of product that can be formed.
1. Convert the given masses of ZnS and O2 to moles:
Moles of ZnS = mass of ZnS / molar mass of ZnS
= 12.85 g / 97.474 g/mol
≈ 0.1319 mol
Moles of O2 = mass of O2 / molar mass of O2
= 15.91 g / 31.9988 g/mol
≈ 0.4972 mol
2. Determine the mole ratio of ZnS to ZnO from the balanced chemical equation:
ZnS + O2 → ZnO + SO2
From the equation, we see that the mole ratio of ZnS to ZnO is 1:1. Therefore, 0.1319 mol of ZnS will form 0.1319 mol of ZnO.
3. Convert the moles of ZnO to grams:
Mass of ZnO = moles of ZnO × molar mass of ZnO
= 0.1319 mol × 81.4084 g/mol
≈ 10.74 g
Therefore, the maximum amount of ZnO that can be formed from 12.85 g of ZnS and 15.67 g of O2 is approximately 10.74 grams.