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 95.0 kg of TiCl4 and 10.0 kg of Na

Question

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 95.0 kg of TiCl4 and 10.0 kg of Na

Answer 1

can any one post the equation.

It would help a lot!!
Thank u..

Answer 2

This is a limiting reagent problem. I worked this for someone a day or so ago but I can't find it. Here is a site that gives a worked example of a limiting reagent problem; just follow these steps.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

Answer 3

Why posting the question that only need High School grade to answer? Are you Junior High School student or lower grade? If you have no intention to put some effort in your study from the start, no need to enroll and then cheat.

Answer 4

To calculate the amount of titanium produced in this reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that will be completely consumed, thus determining the maximum amount of product that can be produced.

First, we need to determine the stoichiometry of the reaction by balancing the equation:

2 Na + TiCl4 -> Ti + 2 NaCl

From the balanced equation, we can see that 2 moles of sodium (Na) react with 1 mole of titanium tetrachloride (TiCl4) to produce 1 mole of titanium (Ti) and 2 moles of sodium chloride (NaCl).

Next, we calculate the number of moles for each reactant:

Moles of TiCl4 = mass of TiCl4 / molar mass of TiCl4
= 95.0 kg / (47.88 g/mol + 4 * 35.45 g/mol)
= 95.0 kg / (47.88 + 141.80)
= 95.0 kg / 189.68 g/mol
= 0.5018 mol

Moles of Na = mass of Na / molar mass of Na
= 10.0 kg / 22.99 g/mol
= 10.0 kg / 22.99 g/mol
= 0.4349 mol

Now, we compare the mole ratios of the reactants to determine the limiting reactant:

Mole ratio of TiCl4 to Ti = 1:1
Mole ratio of Na to Ti = 2:1

Since the mole ratio of Na to Ti is 2:1, and we have 0.4349 moles of Na, we can calculate the maximum amount of Ti that can be produced:

Moles of Ti = 0.4349 mol Na * (1 mol Ti / 2 mol Na)
= 0.2175 mol Ti

Finally, we calculate the mass of titanium produced:

Mass of Ti = moles of Ti * molar mass of Ti
= 0.2175 mol * 47.88 g/mol
= 10.41 g

Therefore, the amount of titanium produced in this reaction is 10.41 kg.

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 95.0 kg of TiCl4 and 10.0 kg of Na

10.4kg

can any one post the equation.

This is a limiting reagent problem. I worked this for someone a day or so ago but I can't find it. Here is a site that gives a worked example of a limiting reagent problem; just follow these steps.

Why posting the question that only need High School grade to answer? Are you Junior High School student or lower grade? If you have no intention to put some effort in your study from the start, no need to enroll and then cheat.

To calculate the amount of titanium produced in this reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that will be completely consumed, thus determining the maximum amount of product that can be produced.