Answer
1. There is a total of 0.0741 mole of $S_2Cl_2$ in 10.0 g.
2. There is a total of 0.128 mole of $C_6H_6$ in 10.0 g.
Work Step by Step
1. As we have determined in 7.71b, the molar mass for $S_2Cl_2$ is equal to 135.04 g/mole. Therefore, the conversion factors are:
$ \frac{1 \space mole \space (S_2Cl_2)}{ 135.04 \space g \space (S_2Cl_2)}$ and $ \frac{ 135.04 \space g \space (S_2Cl_2)}{1 \space mole \space (S_2Cl_2)}$
2. Calculate the number of moles $(S_2Cl_2)$
$ 10.0g \times \frac{1 mole}{ 135.04g} = 0.0741 mole$
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3. As we have determined in 7.71b, the molar mass for $C_6H_6$ is equal to 78.11 g/mole. Therefore, the conversion factors are:
$ \frac{1 \space mole \space (C_6H_6)}{ 78.11 \space g \space (C_6H_6)}$ and $ \frac{ 78.11\space g \space (C_6H_6)}{1 \space mole \space (C_6H_6)}$
4. Calculate the number of moles $(C_6H_6)$
$ 10.0g \times \frac{1 mole}{ 78.11g} = 0.128 moles$
