Tin (Sn) exists in Earth's crust as SnO2. Calculate the percent composition by mass of Sn and O in SnO2.
2 answers:
Answer:
Sn = 78.77%
O = 21.23%
Explanation:
The atomic mass of Sn = 118.71 g/mol
The atomic mass of O = 16 g/mol
The molar mass of SnO₂ = atomic mass of Sn + 2 X atomic mass of O
The molar mass of SnO₂= 150.71
Now, in each mole of SnO₂ there is one mole of Sn and two moles of O
The mass of Sn in 150.71 g of SnO₂ = 118.71
Percentage of Sn in 150.71 g of SnO₂ =
% of Sn = %.
Rest will be O = 100-78.77 = 21.23%
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Answer:
The correct option is;
B) 179 g
Explanation:
The parameters given are;
Mass of H₂ that takes part in the reaction = 2.23 g
Molar mass of hydrogen gas, H₂ = 2.016 g
Number of moles, n, of hydrogen gas H₂ is given by the relation;
Chemical equation for the reaction;
H₂ + Br₂ → 2HBr
Given that one mole of H₂ reacts with one mole of Br₂ to produce two moles of HBr
1.106 mole of H₂ will react with 1.106 mole of Br₂ to produce 2 × 1.106 which is 2.212 moles of HBr
The molar mass, of HBr = 80.91 g/mol
The mass of HBr produced = Molar mass of HBr × Number of moles of HBr
The mass of HBr produced = 80.91 × 2.212 = 178.997 g ≈ 179 grams
Therefore, the correct option is B) 179 g.
20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO =
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO