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Answer:
Mass of NH₃ produced = 34 g
Explanation:
Given data:
Mass of nitrogen = 28 g
Mass of Hydrogen = 12 g
Mass of NH₃ produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 28 g/ 28 g/mol
Number of moles = 1 mol
Moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 12 g/ 2 g/mol
Number of moles = 6 mol
Now we will compare the moles of hydrogen and nitrogen with ammonia.
H₂ : NH₃
3 : 2
6 : 2/3×6 = 4 mol
N₂ : NH₃
1 : 2
Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.
Mass of ammonia produced:
Mass = number of moles × molar mass
Mass = 2 mol × 17 g/mol
Mass = 34 g
A contains 38.5 g of tin for each 12.3 g of fluorine:
<span>mole ratio: </span>
<span>(38.5 g)/(118.71 g/mol):(12.3 g)/(18.998 g/mol) = 0.324:0.647 = 1:2 ⇒ SnF₂ </span>
<span>B contains 56.5 g of tin for each 36.2 g of fluorine: </span>
<span>mole ratio: </span>
<span>(56.5 g)/(118.71 g/mol):(36.2 g)/(18.998 g/mol) = 0.476:1.905 = 1:4 ⇒ SnF₄
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Covalent since they are both non metals, a metal and a non metal make an ionic bond
