Question

oduct (in amounts of
part b
2.2 g ti, 1.8 g f₂
express your answer using two significant figures.
m = g
submit request answer
part c

Explanation:

Step1: Write the balanced reaction

The reaction between Ti and \( F_2 \) is \( \text{Ti} + 2\text{F}_2 ightarrow \text{TiF}_4 \)

Step2: Calculate moles of Ti

Molar mass of Ti \( = 47.87 \, \text{g/mol} \)
Moles of Ti \( = \frac{2.2 \, \text{g}}{47.87 \, \text{g/mol}} \approx 0.04596 \, \text{mol} \)

Step3: Calculate moles of \( F_2 \)

Molar mass of \( F_2 = 38.00 \, \text{g/mol} \)
Moles of \( F_2 = \frac{1.8 \, \text{g}}{38.00 \, \text{g/mol}} \approx 0.04737 \, \text{mol} \)

Step4: Determine limiting reactant

From the reaction, 1 mol Ti reacts with 2 mol \( F_2 \)
Moles of \( F_2 \) required for 0.04596 mol Ti \( = 2 \times 0.04596 = 0.09192 \, \text{mol} \)
But we have only 0.04737 mol \( F_2 \), so \( F_2 \) is limiting.

Step5: Moles of \( \text{TiF}_4 \) formed

From reaction, 2 mol \( F_2 \) forms 1 mol \( \text{TiF}_4 \)
Moles of \( \text{TiF}_4 = \frac{0.04737 \, \text{mol} \, F_2}{2} = 0.023685 \, \text{mol} \)

Step6: Mass of \( \text{TiF}_4 \)

Molar mass of \( \text{TiF}_4 = 47.87 + 4 \times 19.00 = 123.87 \, \text{g/mol} \)
Mass \( = 0.023685 \, \text{mol} \times 123.87 \, \text{g/mol} \approx 2.93 \, \text{g} \) (before rounding)
With two significant figures, mass \( \approx 2.9 \, \text{g} \) (Wait, recalculate: Wait, moles of \( F_2 \) is 0.04737, divided by 2 is 0.023685. Multiply by 123.87: 0.023685*123.87 ≈ 2.93. But let's check again. Wait, maybe I made a mistake in limiting reactant. Wait, Ti: 0.04596 mol, F2: 0.04737 mol. The ratio of F2 to Ti needed is 2:1. So for Ti, F2 needed is 0.04596*2=0.0919, but we have 0.04737, so F2 is limiting. So moles of TiF4 is 0.04737/2=0.023685. Mass: 0.023685*123.87 ≈ 2.93. Now, with two significant figures, 2.9 g? Wait, no, 1.8 has two sig figs, 2.2 has two. Wait, maybe my calculation of moles of F2: 1.8/38 = 0.047368. Then moles of TiF4: 0.047368/2 = 0.023684. Mass: 0.023684*123.87 = 0.023684*123.87 ≈ 2.93. Rounding to two significant figures: 2.9 g? Wait, no, 2.93 is approximately 2.9 when rounded to two significant figures? Wait, 2.93, the first two sig figs are 2 and 9, the next digit is 3, which is less than 5, so it stays 2.9? Wait, no, 2.93 is 2.9 when rounded to two significant figures? Wait, no, 2.93: the first significant figure is 2, second is 9, third is 3. So rounding to two, it's 2.9? Wait, but let's check again. Wait, maybe I messed up the limiting reactant. Wait, moles of Ti: 2.2/47.87 ≈ 0.04596. Moles of F2: 1.8/38 ≈ 0.04737. The stoichiometric ratio of Ti to F2 is 1:2. So for Ti, F2 required is 0.04596*2=0.0919, which is more than available F2 (0.04737). So F2 is limiting. So moles of TiF4 is (0.04737)/2 = 0.023685. Mass: 0.023685*123.87 = 2.93 g. Now, 2.93 with two significant figures: look at the first two digits: 2 and 9, the next digit is 3, so we round down, so 2.9 g? Wait, but 2.93 is closer to 2.9 or 3.0? Wait, 2.93: the third digit is 3, which is less than 5, so we keep the second digit as 9, so 2.9. Wait, but let's check the calculation again. Wait, maybe I made a mistake in molar mass of TiF4. Ti is 47.87, F is 19.00, so 4*19=76, 47.87+76=123.87, correct. Moles of F2: 1.8/38=0.047368, correct. Moles of TiF4: 0.047368/2=0.023684, correct. Mass: 0.023684*123.87=2.93, correct. So with two significant figures, 2.9 g? Wait, but 1.8 and 2.2 have two significant figures, so the answer should have two. So 2.9 g? Wait, no, 2.93 is 2.9 when rounded to two significant figures? Wait, 2.93: the first significant figure is 2 (non-zero), second is 9, third is 3. So we round to two, so 2.9. Alternatively, maybe I made a mistake in limiting reactant. Wait, what if Ti is limiting? Wait, moles of F2 available is 0.04737, moles of Ti that can react with F2: 0.04737/2=0.023685. Which is less than moles of Ti (0.04596). So yes, F2 is limiting, so Ti is in excess. So the calculation is correct. So mass is 2.9 g? Wait, but let's check with another approach. Let's calculate mass of TiF4 from F2. Moles of F2: 1.8/38=0.04737. Moles of TiF4: 0.04737/2=0.023685. Mass: 0.023685*123.87=2.93, which is 2.9 g when rounded to two significant figures.

Answer:

\( \boxed{2.9} \)