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Answer:
15.2 grams of calcium chloride are produced and HCl is the limiting reactant.
Explanation:
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In this case, according to the described scenario, it is possible to realize that the reaction between hydrochloric acid and calcium hydroxide is:
Whereas there is a 2:1 mole ratio of the acid to the base. In such a way, with the given masses, we can compute how much calcium chloride product is produced due to the chemical reaction via stoichiometry:
Whereas we infer that the correct amount is 15.2 g since HCl is the limiting reactant as it produces the fewest grams of the desired product. Consequently, the calcium hydroxide is the excess reactant here.
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Explanation:
Given parameters:
Mass of He = 5.5g
Mass of Ne = 15g
Mass of Kr = 35g
Pressure at STP = 1atm
Unknown:
Partial pressure per gas = ?
Solution:
Partial pressure is the pressure a gas would exert it alone occupies the container.
Partial pressure of a gas = Mole fraction of a gas x Total pressure of mixture
Number of moles =
Number of moles of He = = 1.375mole
Number of moles of Ne = = 0.75mole
Number of moles of Kr = = 0.42mole
Sum total of moles = 1.375 + 0.75 + 0.42 = 2.54moles
Partial pressure of He = x 1atm = 0.54atm
Partial pressure of Ne = x 1atm = 0.29atm
Partial pressure of kr = x 1atm = 0.17atm
Answer:
Explanation:
Mass of water obtained = 42.8 mg
1 mg = 10⁻³ g
Moles of = 42.8/18 = 2.3778×10⁻³ moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
Moles of H = 2 x 2.4 = 4.7556×10⁻³ moles
Molar mass of H atom = 1.008 g/mol
Mass of H in molecule = 4.7556×10⁻³ x 1.008 = 4.7936×10⁻³ g
Given that the cumene only contains hydrogen and carbon. So,
Mass of C in the sample = Total mass - Mass of H
Mass of the sample = 47.6 mg = 47.6×10⁻³ g
Mass of C in sample = 47.6×10⁻³ - 4.80×10⁻³ = 42.8064×10⁻³ g
Moles of C = 42.8064×10⁻³ / 12 = 3.5672×10⁻³ moles
Taking the simplest ratio for H and C as:
4.7556×10⁻³ : 3.5672×10⁻³ = 4 : 3
The empirical formula is =
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 3×12 + 4×1 = 40 g/mol
115 g/mol < Molar mass > 125 g/mol
So,
Molecular mass = n × Empirical mass
115 g/mol < 40 n > 125 g/mol
⇒ n ≅ 3
The formula of cumene =