Question

Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen

Answer 1

11.3 g.

Step-by-step explanation:

Hello there!

In this case, since the combustion of butane is:

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

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

`m_(H_2O)=7.26gC_4H_(10)*(1molC_4H_(10))/(58.14gC_4H_(10))*(5molH_2O)/(1molC_4H_(10)) *(18.02gH_2O)/(1molH_2O)`

Therefore, the resulting mass of water is:

`m_(H_2O)=11.3gH_2O`

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