## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 4 - Exercises - Page 188: 48

#### Answer

7.45 g of $P_4$

#### Work Step by Step

1. Identify the limiting reactant. - Calculate or find the molar mass for $P_4$: $P_4$ : ( 30.97 $\times$ 4 )= 123.88 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$45.69 \space g \times \frac{1 \space mole}{ 123.88 \space g} = 0.3688 \space mole$$ - Calculate or find the molar mass for $Cl_2$: $Cl_2$ : ( 35.45 $\times$ 2 )= 70.90 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$131.3 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 1.852 \space moles$$ Find the amount of product if each reactant is completely consumed. $$0.3688 \space mole \space P_4 \times \frac{ 4 \space moles \ PCl_3 }{ 1 \space mole \space P_4 } = 1.475 \space moles \space PCl_3$$ $$1.852 \space moles \space Cl_2 \times \frac{ 4 \space moles \ PCl_3 }{ 6 \space moles \space Cl_2 } = 1.235 \space moles \space PCl_3$$ Since the reaction of $Cl_2$ produces less $PCl_3$ for these quantities, it is the limiting reactant. 2. Find the amount of $P_4$ consumed: $$1.852 \space moles \space Cl_2 \times \frac{ 1 \space mol \space P_4 }{ 6 \space mol \space Cl_2 } = 0.3087 \space mol \space P_4$$ 3. Subtract that amount from the amount in the reaction mixture. $$Excess = 0.3688 - 0.3087 = 0.0601 \space mol \space P_4$$ 4. Using its molar mass, find the amount in grams: $$0.0601 \space mole \times \frac{ 123.88 \space g}{1 \space mole} = 7.45 \space g$$

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