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if 3.00 g g of titanium metal is reacted with 6.00 g g of chlorine gas, cl2 c l 2 , to form 7.7 g g of titanium (iv) chloride in a combination reaction, what is the percent yield of the product?

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The percentage yield of titanium IV chloride is 95.8%.

The equation of the reaction is;

Ti(s) + 2Cl2(g) ------> TiCl4(s)

Number of moles of Ti = 3.00 g /48 g/mol = 0.0625 moles

Number of moles of Cl2 = 6.00 g/71 g/mol = 0.0845 moles

The limiting reactant yields the least amount of product.

For Ti, the reaction is 1:1 so it yields 0.0625 moles of TiCl4

For Cl2;

2 moles of Cl2 yields 1 mole of TiCl4

0.0845 moles of Cl2 yields 0.0845 moles × 1 mole/2 moles = 0.0423 moles of TiCl4

Hence, Cl2 is the limiting reactant.

Mass of TiCl4 produced = 0.0423 moles × 190 g/mol = 8.037 g

Note that 190 g/mol is the molar mass of TiCl2 and 8.037 g is the theoretical yield of the compound.

Applying the formula;

%yield = actual yield/theoretical yield × 100/1

In the question, the actual yield was supplied as  7.7 g.

%yield =7.7 g/8.037 g × 100/1

%yield = 95.8%

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