Combine 0.2 g atom of silicon with 21.3 g of chlorine. Find the empirical formula of the compound formed
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HI...here is ur answer
Si+2Cl2⟶SiCl4Si+2Cl2⟶SiCl4
Molecular mass of Si=28, Cl=35.5
Initially we will have to check if any chemical is present in excess:
So, 28 g of Silicon will react with 142 g of chlorine
Hence 0.2 g of Silicon will react with 14228×0.214228×0.2 g= 1.014 g of chlorine.
Hence Silicon is the limiting reagent (as soon as it is finished, the reaction is stopped)
The product formed is SiCl4. We will reject the excess chlorine since it is not a “formed product”.
Now, Molar mass of SiCl4=28+35.5×4=17028+35.5×4=170
So, 0.2 g silicon gives 17028×0.217028×0.2 g= 1.214 g of SiCl4
% of Silicon=0.21.2140.21.214=0.1648=16.48%
% of Chlorine= 1–0.1648=83.52%
THIS WILL HELP U
Si+2Cl2⟶SiCl4Si+2Cl2⟶SiCl4
Molecular mass of Si=28, Cl=35.5
Initially we will have to check if any chemical is present in excess:
So, 28 g of Silicon will react with 142 g of chlorine
Hence 0.2 g of Silicon will react with 14228×0.214228×0.2 g= 1.014 g of chlorine.
Hence Silicon is the limiting reagent (as soon as it is finished, the reaction is stopped)
The product formed is SiCl4. We will reject the excess chlorine since it is not a “formed product”.
Now, Molar mass of SiCl4=28+35.5×4=17028+35.5×4=170
So, 0.2 g silicon gives 17028×0.217028×0.2 g= 1.214 g of SiCl4
% of Silicon=0.21.2140.21.214=0.1648=16.48%
% of Chlorine= 1–0.1648=83.52%
THIS WILL HELP U
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