Answer
a. There should be used 0.076 g of $C_6H_8O_7$ for every $1.0 \times 10^2 \space mg \space NaHCO_3$
b. 0.052 g of $CO_2$ could be produced.
Work Step by Step
a.
$ NaHCO_3 $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 84.01 g/mol
$$ \frac{1 \space mole \space NaHCO_3 }{ 84.01 \space g \space NaHCO_3 } \space and \space \frac{ 84.01 \space g \space NaHCO_3 }{1 \space mole \space NaHCO_3 }$$
$ C_6H_8O_7 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 7 )= 192.12 g/mol
$$ \frac{1 \space mole \space C_6H_8O_7 }{ 192.12 \space g \space C_6H_8O_7 } \space and \space \frac{ 192.12 \space g \space C_6H_8O_7 }{1 \space mole \space C_6H_8O_7 }$$
$$ 1.0 \times 10^{2} \space mg \space NaHCO_3 \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mole \space NaHCO_3 }{ 84.01 \space g \space NaHCO_3 } \times \frac{ 1 \space mole \space C_6H_8O_7 }{ 3 \space moles \space NaHCO_3 } \times \frac{ 192.12 \space g \space C_6H_8O_7 }{1 \space mole \space C_6H_8O_7 } = 0.076 \space g \space C_6H_8O_7 $$
b.
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$$ \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$
$$1.0 \times 10^{2} \space mg \space NaHCO_3 \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mole \space NaHCO_3 }{ 84.01 \space g \space NaHCO_3 } \times \frac{ 3 \space moles \space CO_2 }{ 3 \space moles \space NaHCO_3 } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 0.052 \space g \space CO_2 $$
