42 g of 95% pure MgCO₃ is required to neutralize 2 L of semi normal HCl.
Let's consider the balanced equation for the reaction between magnesium carbonate and hydrochloric acid.
MgCO₃ + 2 HCl ⇒ MgCl₂ + H₂O + CO₂
We will calculate the moles in 2 L of 0.5 N HCl, taking into account that there is 1 equivalent per mole of HCl.
[tex]2L \times \frac{0,5eq}{L} \times \frac{1mol}{eq} = 1 mol[/tex]
The moles of MgCO₃ that react with 1 mol of HCl are:
[tex]1molHCl \times \frac{1molMgCO_3}{2molHCl} = 0.5 mol MgCO_3[/tex]
The molar mass of MgCO₃ is 84.31 g/mol. The mass corresponding to 0.5 moles of MgCO₃ is:
[tex]0.5 mol \times \frac{84.31g}{mol} = 40 g[/tex]
There are 95 g of pure MgCO₃ every 100 g of impure MgCO₃. The mass of impure MgCO₃ that contains 40 g of pure MgCO₃ is:
[tex]40gPure \times \frac{100gImpure}{95gPure} \approx 42 g Impure[/tex]
42 g of 95% pure MgCO₃ is required to neutralize 2 L of semi normal HCl.
You can learn more about neutralization here: https://brainly.com/question/16188332
