A vessel contains 8 g of TiO2- 2-4 g of carbon and 28-4 g of Cl2 gas- Find the maximum mass of TiCl4 which can be produced-3TiO2s - 4Cs - 6Cl2g - 3TiCl4g - 2CO2g - 2COgConsider the reaction to go to completion- atomic mass of Ti - 48 u- C - 12 u- O - 16 u- Cl - 35-5 u

The balanced chemical equation for above mentioned reaction is: 3TiO_2(s) + 4C(s) + 6Cl_2(g) → 3TiCl_4(g) + 2CO_2(g) + 2CO(g) Number of moles of TiO_2 : 880 = 0 ...

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Byju's Answer

Standard XIII

Chemistry

Stoichiometric Calculations

A vessel cont...

Question

A vessel contains 8 g of $T i O_{2}$ , 2.4 g of carbon and 28.4 g of $C l_{2}$ gas. Find the maximum mass of $T i C l_{4}$ which can be produced.
$3 T i O_{2} (s) + 4 C (s) + 6 C l_{2} (g) \to 3 T i C l_{4} (g) + 2 C O_{2} (g) + 2 C O (g)$

(Consider the reaction to go to completion; atomic mass of $T i = 48 u, C = 12 u, O = 16 u, C l = 35.5 u$ )

1.9 g

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28 g

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19 g

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3.8 g

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Solution

The correct option is C 19 g
The balanced chemical equation for above mentioned reaction is:
$3 T i O_{2} (s) + 4 C (s) + 6 C l_{2} (g) \to 3 T i C l_{4} (g) + 2 C O_{2} (g) + 2 C O (g)$
Number of moles of $T i O_{2} : \frac{8}{80} = 0.1$ mol $T i O_{2}$
Number of moles of Carbon $= \frac{2.4}{12} = 0.2$ mol $C$
Number of moles of Chlorine $= \frac{28.4}{71} = 0.4$ mol $C l_{2}$
To find limiting reagent, we calculate $\frac{Given moles}{Stoichiometric coefficient}$
For $T i O_{2} = \frac{0.1}{3} = 0.033$
For Carbon $= \frac{0.2}{4} = 0.05$
For Chlorine $= \frac{0.4}{6} = 0.06$
Since, the value for $T i O_{2}$ is minimum, it will be the limiting reagent.
As per stoichiometry,
3 moles of $T i O_{2}$ reacts with 4 moles of $C$ and 6 moles of $C l_{2}$ to give 3 moles of $T i C l_{4}$ .
0.1 mol of $T i O_{2}$ will produce 0.1 mol of $T i C l_{4}$
Mass of $T i C l_{4}$ produced = 0.1 mol $\times$ molar mass = 0.1 mol $\times$ 190 g/mol
Mass of $T i C l_{4}$ produced = 19 g of $T i C l_{4}$

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A vessel contains 8 g of TiO2, 2.4 g of carbon and 28.4 g of Cl2 gas. Find the maximum mass of TiCl4 which can be produced. 3TiO2(s)+4C(s)+6Cl2(g)→3TiCl4(g)+2CO2(g)+2CO(g) (Consider the reaction to go to completion; atomic mass of Ti=48 u, C=12 u, O=16 u, Cl=35.5 u)