Methane and hydrogen sulfide form when hydrogen reacts with carbon disulfide. identify the excess reagent and calculate how much remains after 36 l of h2reacts with 12 l of cs2.

3 Litres.

Explanation:

The balance equation of reaction is written in equation (1) below;

4H2 (g) + CS2(g) > CH4 (g) + 2H2S (g) (1).

(From the balanced equation above) one mole of methane, CH4 and two moles of Hydrogen Sulphide, H2S are being produced when one mole of carbon Sulphide, CS2 reacts with 4 moles of Hydrogen gas.

(Parameters given from the question): we have 36 litres of Hydrogen, H2 gas and 12 litres of carbon Sulphide, CS2 gas.

Therefore, 36 Litres of H2 would react completely with; 36 litres/ 4.

= 9 Litres of Carbon Sulphide.

LIMITING REAGENT: (The reagent that is totally consumed). Hydrogen gas, H2 is the limiting reactant since we have 12 litres of Carbon Sulphide, CS2.

EXCESS REAGENT: Carbon Sulphide.

Amount remaining= 12 Litres of CS2 - 9 litres that was reacted.

= 3 Litres.

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

the atomic mass is the bottom number on the element.

-steel jelly