# Chapter 4 - Exercises - Page 187: 31

9.3 g of $HBr$ are needed to dissolve 3.2 g of pure iron. This reaction produces 0.12 g of $H_2$

#### Work Step by Step

1. $Fe$ : 55.85 g/mol $$\frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \space and \space \frac{ 55.85 \space g \space Fe }{1 \space mole \space Fe }$$ $HBr$ : ( 79.90 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 80.91 g/mol $$\frac{1 \space mole \space HBr }{ 80.91 \space g \space HBr } \space and \space \frac{ 80.91 \space g \space HBr }{1 \space mole \space HBr }$$ $$3.2 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \times \frac{ 2 \space moles \space HBr }{ 1 \space mole \space Fe } \times \frac{ 80.91 \space g \space HBr }{1 \space mole \space HBr } = 9.3 \space g \space HBr$$ 2. $$\frac{1 \space mole \space H_2 }{ 2.016 \space g \space H_2 } \space and \space \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 }$$ $$3.2 \space g \space Fe \times \frac{1 \space mole \space Fe }{ 55.85 \space g \space Fe } \times \frac{ 1 \space mole \space H_2 }{ 1 \space mole \space Fe } \times \frac{ 2.016 \space g \space H_2 }{1 \space mole \space H_2 } = 0.12 \space g \space H_2$$

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