What volume of hydrogen is obtained from the reaction 4.8g of magnesium metal with excess hydrochloric acid? ` Mg(s) +2HCl(aq) -> MgCl_2(aq)+H_2` Please explain. Thank you.
4.8 g of magnesium metal reacts with excess hydrochloric acid and the chemical equation of the same is:
`Mg(s) + 2HCl(aq) -> MgCl_2 + H_2`
One mole of magnesium reacts with 2 moles of hydrochloric acid to give one mole of hydrogen gas.
The molar mass of magnesium is 24, 4.8 g of magnesium is equivalent to 4.8/24 = 0.2 moles. When this reacts with the acid 0.2 mole of hydrogen gas is released.
At STP, the volume of 1 mole of an ideal gas is 22.4 liter. The volume of hydrogen can be approximated to that of an ideal gas. The volume of 0.2 mole of hydrogen is 4.48 liters.
The volume of hydrogen is obtained from the reaction 4.8g of magnesium metal with excess hydrochloric acid at STP is 4.48 liters.