Calculate the mass of water produced when 5.83 g of butane reacts with excess oxygen. express your answer to three significant figures and include the appropriate units.

First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
9.04 g of H2O