Question

What volume of oxygen measured at s.t.p would be required for complete combustion of all hydrogen evolved by reacting 12g of magnesium with excess tetraoxosulphate(vi) acid

Answer 1

Here,tetraoxosulphate(vi) acid is sulphuric acid whose reaction with magnesium is

$Mg+H_2SO_4\longrightarrow MgSO_4+H_2(g)$

Here sulphuric acid is in excess whereas Mg mass is given as 12 g

moles of Magnesium is = $\frac{given mass}{molar mass}=\frac{12}{24}=0.5 mole$

since from above reaction 1 mole of Magnesium givesn 1 mole of hydrogen so, moles of

hydrogen gas produced= 0.5 mole

Now, it is combustion with oxygen so the reaction will be as follows

$2H_2+O_2\longrightarrow 2H_2O$

From above reaction we can say that 2 mole of hydrogen react 1 mole oxygen

so 0.5 mole of hydrogen will react with 0.25 mole of Oxygen.

We know that at STP 1 mole of gas has volume = 22.4 L

so, volume of oxygen require= $22.4\times0.25=5.6 L$

So, volume of oxygen required = 5.6 L