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Calculate the mass of water produced when 1.57g of butane reacts with excess oxygen

Sagot :

Explanation:

So, first you will want to write the balanced chemical equation for this reaction.

Butane = [tex]C_4H_{10}[/tex]

[tex]2C_4H_{10}+13O_2=>10H_2O+8CO_2[/tex]

^ This ends up being your balanced chemical equation. Now, you can do the math!

[tex]1.57gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}}*\frac{10molH_2O}{2molC_4H_{10}}*\frac{18gH_2O}{1molH_2O}[/tex]

After plugging this into a calculator, your final mass of water should be:

2.43gH2O