Final answer:
Using the theoretical mass (197 g from 1.00 kg of titanium tetrachloride), assuming the entire mixture produces titanium at the same rate, the calculation leads to an approximate yield of 219.46 g of titanium.
Step-by-step explanation:
If the second reaction is carried out with an 85.7% yield, to determine the mass of titanium that can be obtained from 1.30 kg of the ilmenite-sand mixture, we need to assess the maximum possible yield first. In an earlier step, multiplying the number of moles of titanium (4.12 mol) by the molar mass of titanium (47.867 g/mol) provided the theoretical mass of titanium obtainable, which is 197 g from 1.00 kg of titanium tetrachloride.
Now, we apply the actual yield percentage to the theoretical yield to get the actual mass of titanium produced:
- Theoretical mass of Ti: 197 g
- Actual yield: 85.7%
The actual mass of titanium produced = Theoretical mass × percent yield/100
The calculation goes as follows:
197 g × 85.7/100 = 168.839 g of Ti (rounded to 168.84 g)
Since 1.00 kg of titanium tetrachloride produces 197 g of Ti, 1.30 kg of the mixture will proportionally yield more Ti, following the same percentage of actual yield (85.7%).
The exact mass would depend on the titanium content of the ilmenite-sand mixture. However, assuming that all of the mixture is used and produces titanium at the same rate:
1.30 kg = 1300 g
1300 g of the mixture would theoretically produce (1300 g / 1000 g/kg) × 197 g = 256.1 g of Ti
Applying the actual yield:
256.1 g × 85.7/100 = 219.4647 g of Ti
Thus, with an 85.7% yield, from 1.30 kg of the ilmenite-sand mixture, one can obtain approximately 219.46 g of titanium.