Given the balanced equation below and knowing that 0.0470 grams of magnesium metal fully reacted, determine how many grams of hydrogen gas formed as a result of that reaction.
Mg (s) + 2 HCl (aq) = MgCl2 (aq) + H2 (g)
The correct answer and explanation is :
Solution:
- Balanced Chemical Equation: Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg} (s) + 2 \text{HCl} (aq) \rightarrow \text{MgCl}_2 (aq) + \text{H}_2 (g)
- Find the Molar Masses:
- Magnesium (Mg) = 24.305 g/mol
- Hydrogen gas (H₂) = 2.016 g/mol
- Convert Mass of Mg to Moles: Moles of Mg=0.0470 g24.305 g/mol\text{Moles of Mg} = \frac{0.0470 \text{ g}}{24.305 \text{ g/mol}} =0.001934 moles of Mg= 0.001934 \text{ moles of Mg}
- Use Stoichiometry to Find Moles of H₂:
- From the balanced equation, 1 mole of Mg produces 1 mole of H₂.
- Therefore, moles of H₂ produced: 0.001934 moles of H₂0.001934 \text{ moles of H₂}
- Convert Moles of H₂ to Mass: Mass of H2=0.001934×2.016 g/mol\text{Mass of H}_2 = 0.001934 \times 2.016 \text{ g/mol} =0.00390 g of H2= 0.00390 \text{ g of H}_2
Final Answer: 0.00390 g of H₂
Explanation:
This reaction is a classic example of a single replacement reaction, where magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Since magnesium is more reactive than hydrogen, it displaces hydrogen from the acid.
To determine the amount of hydrogen gas formed, we used stoichiometry:
- Converted given Mg mass to moles using its molar mass.
- Used the molar ratio (1:1) from the balanced equation to find moles of H₂ produced.
- Converted moles of H₂ to grams using the molar mass of hydrogen gas.
Thus, when 0.0470 g of Mg fully reacts, it produces 0.00390 g of H₂ gas.