(a) Mg+NO3−→Mg2++NH3 The complete balanced equation of the given . reaction is 4‌Mg+NO3−+9H⊕→4Mg2++NH3+3H2O Given, conc. of NO3−=0.1M Volume of NO3−=100‌mL We know that, Molarity =‌
‌ Number of moles ‌
‌ Volume of solution (in L) ‌
‌‌ Hence, number of moles ‌=‌ molarity ‌×‌ volume of solution (in L) ‌ ‌‌ So, number of moles ‌=0.1×100×10−3‌=10m‌mol‌=10−2‌mol From above Eq., 1 mole NO3−reacts with 4 moles Mg. So, IO−2 mole NO3−reacts with 4×10−2 moles Mg. Now, to calculate the mass of Mg required, use the formula, Number of moles =‌
‌ Given mass ‌
‌ Molar mass ‌
Molar mass of Mg=24gmol−1 On substituting the value,‌‌4×10−2=‌
‌ Mass of ‌Mg
24
We will get, ‌‌ Mass of ‌Mg=24×4×10−2 ‌‌ Mass of ‌Mg=0.96g