## Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.

# Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Limiting Reactant - Page 108: 35

#### Answer

(a) Some C is left over. (b) 23 g of C

#### Work Step by Step

1. Balance the reaction: $$SiO_2 + 3 C \longrightarrow SiC + 2CO$$ - Calculate or find the molar mass for $SiO_2$: $SiO_2$ : ( 28.09 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 60.09 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$300. \space g \times \frac{1 \space mole}{ 60.09 \space g} = 4.99 \space moles$$ - Calculate or find the molar mass for $C$: $C$ : 12.01 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$203 \space g \times \frac{1 \space mole}{ 12.01 \space g} = 16.9 \space moles$$ Find the amount of product if each reactant is completely consumed. $$4.99 \space moles \space SiO_2 \times \frac{ 1 \space mole \ SiC }{ 1 \space mole \space SiO_2 } = 4.99 \space moles \space SiC$$ $$16.9 \space moles \space C \times \frac{ 1 \space mole \ SiC }{ 3 \space moles \space C } = 5.63 \space moles \space SiC$$ Since the reaction of $SiO_2$ produces less $SiC$ for these quantities, it is the limiting reactant. $$4.99 \space moles \space SiO_2 \times \frac{ 3 \space moles \ C }{ 1 \space mole \space SiO_2 } = 15.0 \space moles \space C$$ - Using the molar mass as a conversion factor, find the mass in g: $$15.0 \space mole \times \frac{ 12.01 \space g}{1 \space mole} = 180. \space g$$ Excess = Initial - Consumed = 203 g - 180. g = 23 g

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