Question

Calculate the mass of water produced when 1.43 g of butane reacts with excess oxygen. Express your answer to three significant figures and include the appropriate units. g

Answer 1

`m_(H_2O)=2.22gH_2O`

Step-by-step explanation:

Hello!

In this case, since the combustion of butane is:

`C_4H_(10)+(13)/(2) O_2\rightarrow 4CO_2+5H_2O`

As there is an excess of oxygen, we can compute the mass of water by simply using the molar masses of butane and water and the 1:5 mole ratio between them as shown below:

`m_(H_2O)=1.43gC_4H_(10)*(1molC_4H_(10))/(58.14gC_4H_(10)) *(5molH_2O)/(1molC_4H_(10)) *(18.02gH_2O)/(1molH_2O) \\\\m_(H_2O)=2.22gH_2O`

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