Titanium (Ti) can be produced by the reaction of metallic sodium (Na) with titanium tetrachloride vapor (TiCl4). The byproduct of this reaction is sodium chloride (NaCl). Calculate the amount of titanium produced (in kg) when a reactor is charged with 87.0 kg of TiCl4 and 20.0 kg of Na.
you have an answer on sulfur?
the formula.
(Kg of Na)/ 23*4 =X
X * 47.88 = in g
convert to Kg
10.4
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To solve this problem, we need to determine the limiting reagent, which is the reactant that will be completely consumed first, thus determining the maximum amount of product that can be formed. The other reactant is said to be in excess.
Let's start by calculating the amount of titanium that can be produced from each reactant separately.
1. For TiCl4:
The balanced chemical equation for the reaction is:
2 Na + TiCl4 → Ti + 2 NaCl
From the equation, we can see that 1 mole of TiCl4 produces 1 mole of Ti.
The molar mass of TiCl4 is:
Ti: 47.88 g/mol + (4 × Cl: 35.45 g/mol) = 189.68 g/mol
To convert 87.0 kg of TiCl4 to moles, we can use the molar mass:
Number of moles of TiCl4 = (mass of TiCl4) / (molar mass of TiCl4)
= (87.0 kg) / (189.68 g/mol)
= (87.0 * 1000) g / (189.68 g/mol) [Converting kg to g]
= 459.04 mol
Since the stoichiometry of the reaction is 1:1 for TiCl4 to Ti, the amount of Ti produced from 459.04 moles of TiCl4 is also 459.04 moles.
So, the mass of Ti produced from TiCl4 = (459.04 moles) × (47.88 g/mol) = 22002.91 g = 22.0 kg (rounded to two decimal places)
2. For Na:
The balanced chemical equation shows that 2 moles of Na react with 1 mole of TiCl4 to produce 1 mole of Ti.
The molar mass of Na is 22.99 g/mol.
To calculate the number of moles of Na in 20.0 kg:
Number of moles of Na = (mass of Na) / (molar mass of Na)
= (20.0 kg) / (22.99 g/mol)
= (20.0 * 1000) g / (22.99 g/mol) [Converting kg to g]
= 869.98 mol
Using stoichiometry, we can determine the maximum amount of Ti that can be produced from Na.
Since the stoichiometry of the reaction is 2:1 for Na to TiCl4, we need to adjust the number of moles of Na:
Adjusted moles of Na = (869.98 mol) / 2 = 434.99 mol
So, the mass of Ti produced from Na = (434.99 moles) × (47.88 g/mol) = 20828.71 g = 20.8 kg (rounded to two decimal places)
To find the limiting reagent, we compare the amounts of Ti produced from each reactant. The smaller value corresponds to the limiting reagent.
In this case, the limiting reagent is Na, as it produces the smaller amount of Ti compared to TiCl4.
Therefore, the maximum amount of titanium that can be produced is 20.8 kg.