Question

Si + 2 SiCl₄ -> Si₃Cl₈

How many grams of SiCl₄ are needed to produce 1.0 kg of Si₃Cl₈, if the reaction gives you only 72% yield?

Answer 1

Balanced equation:
Si + 2 SiCl₄ ====> Si₃Cl₈
Reaction type: synthesis

Mass of Si3Cl8 = 1 Kg = 1000 gm

Molar mass of Si3Cl8 = 367.88 g/mol  

Moles of of Si3Cl8 = 1000 gm / 367.88 g/mol =  2.718 Moles

Moles of SiCl2 theoretically reacted  =  5.436 Moles

Moles of SiCl2 required when it is 72 % yield = 5.436 moles x 100 / 72 = 7.55 Moles

Molar mass of SiCl4 = 169.89 g/mol

Mass of SiCl4 actually required = 7.55 Moles x 169.89 g/mol =  1282.72 g  

Hence 1282.72 g of SiCl4 is required to to produce 1.0 kg of Si₃Cl₈, if the reaction gives you only 72% yield