How get a dilution at 150g/l of H2SO4 at 96%? I've never done any chemistry before and I would like to get 5 liters of H2SO4 at a concentration of $150 [g/l]$.

One liter of $H_2SO_4$ at $96\% $W/W has a density of $1.84 [g/ml]$.

It means my bottle should weight $1840~[g]$ where I have $96/100\cdot1840=1766~[g]$ of pure acid.

5 liters of acid at $150~[g/l]$ means $5~[l]\cdot 150~[g/l] = 750~[g]$

The amount of acid at 96% is:

$$\frac{1840~[g]}{1766~[g]}\cdot 750~[g] = 781 [g] $$ 

Converted in liters I must have

$$\frac{781 [g]}{1840 [g/l]} = 424~[ml]$$

Eventually I need to poor 424 [ml] of $H_2SO_4$ at 96% into $ 5 [l] - 474 [ml] = 4.51~[l]$ of distilled water. Is that right? 

I am wondering of the amount of heat will I get during this dilution. Is that ok to gently poor it (about 1 minute) into distilled water contained into a plastic jerrycan?
First of all: Never add water to sulphuric acid. The heat of the reaction may lead to an explosion. Use safety glasses, safety gloves, and wear protective clothing.

The densities at 20°C are


Water: $\rho_{water} = 0.9982~\mathrm{g~cm^{-3}}$
15% $\ce{H2SO4}$: $\rho_1 = 1.10~\mathrm{g~cm^{-3}}$
96% $\ce{H2SO4}$: $\rho_2 = 1.84~\mathrm{g~cm^{-3}}$


To get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$ with a mass of

$m = \rho_1 \cdot V = 1.10~\mathrm{g~cm^{-3}} \cdot 5000~\mathrm{cm^3} = 5500~\mathrm{g}$

you need

$m = 0.15 \cdot 5500~\mathrm{g} = 825~\mathrm{g}$ pure $\ce{H2SO4}$

These are contained in

$m = \frac{825~\mathrm{g}}{0.96} = 859.4~\mathrm{g}$ of 96% $\ce{H2SO4}$.

i.e.

$V = \frac{859.4~\mathrm{g}}{1.84~\mathrm{g~cm^{-3}}} = 467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$.

Finally, the amount of water you need to pour the acid in is

$m = 5500~\mathrm{g} - 859.4~\mathrm{g} = 4640.6~\mathrm{g}$ water, i.e. $4649.0~\mathrm{cm^3}$

In short: gently pour, while stirring, $467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$ into $4649~\mathrm{cm^3}$ distilled water to get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$. Break the procedure if the temperature gets too high.

Disclaimer: I do not take any responsibility for the correctness of the above calculation.