Answer:
Explanation:
<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>
- atomic mass of C: 12.0107 g/mol
- molar mass of CO₂: 44.01 g/mol
- Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C
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<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>
- atomic mass of H: 1.00784 g/mol
- molar mass of H₂O: 18.01528 g/mol
- proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H
<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>
- Mass of O = mass of pure sample - mass of C - mass of H
- Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O
Round to four decimals: Mass of O = 0.3840 g
<u>4) Mole calculations</u>
Divide the mass in grams of each element by its atomic mass:
- C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
- H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
- O: 0.3840 g / 15.999 g/mol = 0.02400 mol
<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>
- C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
- H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
- O: 0.02400 mol / 0.02400 mol = 1
Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:
25.2 / 34.6 x 100
which would get you 73%
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Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
Answer:
Copper(II) sulphate – sodium hydroxide reaction
The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.