Answer:
Moles of NaCl formed is 6.0 moles
Explanation:
We are given the equation;
2 Na(s) + Cl₂(g) → 2 NaCl(s)
- Moles of Na is 6.0 moles
- Moles of Cl₂ is 4.0 moles
From the reaction;
2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl
In this case;
6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.
Thus, the rate limiting reagent is sodium.
But, 2 moles of sodium reacts to form 2 moles of NaCl
Therefore;
Moles of NaCl = Moles of Na
= 6.0 moles
Thus, moles of sodium chloride produced is 6.0 moles
Answer:
Have only single bonds.
Explanation:
Hello,
In this case, we need to remember that saturation is a state at which a carbon chain contains no insaturations, that is neither double nor triple bonds such case are alkenes and alkynes, but saturations only which are characterized by the presence of single bonds between adjacent carbon atoms. Such is the case of ethane (CH₃-CH₃), propane (CH₃-CH₂-CH₃), butane (CH₃-CH₂-CH₂-CH₃) and so on.
Best regards.
Answer:
Explanation:
The symbol of beryllium is Be. It is located in the second group on the periodic table.
Beryllium is a divalent cation with a charge of +2.
It is capable of losing 2 electrons in order to be like the noble gas He and this is a more stable configuration.
The answer is TiBr4 which is the formula
A) Limiting reactant
You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)
a) Molar ratios:
6 mol HF : 1 mol SiO2 : 1 mol H2SiF6
2) Molar masses:
Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol
=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol
3) convert data in grams to moles
21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2
70.5 g HF / 20 g/mol = 3.525 mol HF
4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.
6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10
=> There is more HF than the needed to react with 0.35mol of SiO2 =>
SiO2 is the limiting reactant (HF is in excess)
b) Mass of excess reactant.
1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted
0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF
2.1 mol HF * 20 g/mol = 42 gram of HF
2) Subtract the quantity that reacted from the original quantity:
70.5 g - 42 g = 28.5 g of HF in excess
c) Theoretical yield of H2SiF6
1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6
Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams
d) % yield
% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
