$$2C_4H_{10} + 13O_2 -> 8CO_2 + 10H_2O$$

First, calculate the number of moles of butane in 58 g:

$$Moles \ C_4H_{10} = \frac{58 g}{58.12 g/mol} = 1.00 mol$$

According to the balanced chemical equation, 2 moles of butane produce 8 moles of carbon dioxide. Therefore, 1 mole of butane will produce:

$$Moles \ CO_2 = 1.00 mol \ C_4H_{10} \times \frac{8 mol \ CO_2}{2 mol \ C_4H_{10}}$$

$$Moles \ CO_2 = 4.00 mol$$

So, when 58 g of butane burn in oxygen, 4.00 moles of carbon dioxide are formed.