Question

1. How many moles of water vapour are formed when 10 litres of butane gas, CH₁ is burned in oxygen at STP?

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Answer 1

Answer: 2.23 moles

Step-by-step explanation:

I am going to say, first, that the formula for butane is not
`CH_(1)`, but that it is actually
`C_(4)H_(10)`.

Using the formula for butane, we can see that the balanced equation for the combustion of butane is
`2C_(4)H_(10)+13O_(2) - > 8CO_(2) + 10H_(2)O`.

We need to remember that, for an ideal gas at STP, the molar volume is 22.4 L/mol, thus we have
`(10 L)/(22.4(L)/(mol) )`=0.446 moles of butane. We can see that the moles of H2O produced is 5 times the moles of butane used, thus, there are 0.446 * 5 or about 2.23 mol of water formed (rounding errors are bound to occur).