This problem is providing the mass of butane (1.89 g) that undergoes combustion in excess oxygen to produce water and carbon dioxide, so the mass of water product is required and found to be 2.93 g water after the calculations.
Stoichiometry:
In chemistry, stoichiometry is used for us to deal with chemical amounts and calculations, based on the balanced chemical equations, molar masses and mole ratios. In such a way, we first write the balanced equation for the combustion of butane as shown below:
[tex]C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O[/tex]
Where we can see a 1:5 mole ratio of butane to water, which means we can write our stoichiometric setup as follows:
[tex]1.89gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}} *\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O} [/tex]
Where the 58.12 g/mol is the molar mass of butane and 18.02 g/mol that of water, so they are used to convert from grams to moles of butane and from moles to grams of water respectively. Hence, the result turns out to be:
[tex]2.93 g H_2O[/tex]
Learn more about mole ratios: https://brainly.com/question/15288923
