Answer:
180 amu
C₆H₁₂O₆
Explanation:
Step 1: Determine the molecular mass of the compound
The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:
M = m/n = 3.06/0.0170 mol = 180 g/mol
Then, the molecular mass is 180 amu.
Step 2: Determine the molar mass of the empirical formula.
M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)
M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol
Step 3: Determine the molecular formula
First, we will determine "n" according to the following expression.
n = molar mass molecular formula / molar mass empirical formula
n = 180 g/mol / 30 g/mol = 6
The molecular formula is:
n × CH₂O = 6 × CH₂O = C₆H₁₂O₆
Answer:
Mass = 2.89 g
Explanation:
Given data:
Mass of NH₄Cl = 8.939 g
Mass of Ca(OH)₂ = 7.48 g
Mass of ammonia produced = ?
Solution:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
Number of moles of NH₄Cl:
Number of moles = mass/molar mass
Number of moles = 8.939 g / 53.5 g/mol
Number of moles = 0.17 mol
Number of moles of Ca(OH)₂ :
Number of moles = mass/molar mass
Number of moles = 7.48 g / 74.1 g/mol
Number of moles = 0.10 mol
Now we will compare the moles of ammonia with both reactant.
NH₄Cl : NH₃
2 : 2
0.17 : 0.17
Ca(OH)₂ : NH₃
1 : 2
0.10 : 2/1×0.10 = 0.2 mol
Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.17 mol × 17 g/mol
Mass = 2.89 g
Answer: Some signs of a chemical change are a change in color and the formation of bubbles. The five conditions of chemical change: color chage, formation of a precipitate, formation of a gas, odor change, temperature change.
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