Assume carbon burns according to following equation:
$$2C(s) + O_2(g) \rightarrow 2CO(s)$$
when 12 g carbon is burnt in 48 g of oxygen, the volume of carbon monoxide produced is ______ $$\times 10^{-1}$$ L at STP [nearest integer]
[Given: Assume CO as ideal gas, Mass of C is 12 g mol$$^{-1}$$, mass of O is 16 g mol$$^{-1}$$ and molar volume of an ideal gas at STP is 22.7 L mol$$^{-1}$$]

We are given the reaction $$2C(s) + O_2(g) \rightarrow 2CO(g)$$ and that 12 g of carbon is burnt in 48 g of oxygen.

To determine the amount of carbon monoxide formed, we first calculate the moles of each reactant. Moles of C = $$\frac{12}{12} = 1$$ mol and moles of O$$_2$$ = $$\frac{48}{32} = 1.5$$ mol.

From the stoichiometric relationship 2 mol C requires 1 mol O$$_2$$, it follows that 1 mol C requires 0.5 mol O$$_2$$. Since there are 1.5 mol O$$_2$$ available, oxygen is in excess and carbon is the limiting reagent.

Because 2 mol C produce 2 mol CO, 1 mol C will produce 1 mol CO, so the amount of CO produced is 1 mol.

At STP, the molar volume is 22.7 L/mol, giving a volume of 1 mol × 22.7 L/mol = 22.7 L, which can be expressed as 227 × 10$$^{-1}$$ L.

The correct answer is 227.