Question

7. Calculate the volume of hydrogen gas liberated
when 0.3 g of magnesium is treated with dilute hydrochloric acid
​

Answer 1

Mg(s)+2HCl(aq)→Mg Cl2(aq)+H2(g)↑⏐. Explanation: Moles of metal, = 4.86⋅g24.305⋅g⋅mol−1 ...

pls mark Brainliest

Answer 2

Answer:

i think the answer is 0.28 litres

Explanation:

1 mole of magnesium reacts with 2 moles of HCL to form 1 mole of hydrogen gas  so 0.3 grams which is 0.0125 mole of magnesium reacts to give 0.0125 moles of hydrogen gas

Now we know that a 1 mole of  gas occupies 22.4 litres at STP now for 0.0125 moles of hydrogen the volume of the gas liberated is 0.28 litres

the equation is like

Mg+ 2 HCl =MgCl₂ + H₂