Question

In the single replacement reaction of solid lithium and aqueous magnesium nitrate, what mass of lithium combines with 75.0 grams of magnesium nitrate? (7.02 g Li) ___Li(s) + ___Mg(NO3)2(aq) → ___Mg(s) + ___ LiNO3(aq)

Answer 1

The balanced chemical equation for the single replacement reaction of solid lithium and aqueous magnesium nitrate is:

2Li(s) + Mg(NO3)2(aq) → Mg(s) + 2LiNO3(aq)

From the equation, we can see that 2 moles of Li react with 1 mole of Mg(NO3)2 to produce 1 mole of Mg and 2 moles of LiNO3.

First, we need to determine the number of moles of Mg(NO3)2 in 75.0 grams:

m = mass / molar mass

m = 75.0 g / 148.31 g/mol

m = 0.5053 mol Mg(NO3)2

Since 2 moles of Li react with 1 mole of Mg(NO3)2, we need half as many moles of Li:

n(Li) = 0.5 x n(Mg(NO3)2)

n(Li) = 0.5 x 0.5053 mol

n(Li) = 0.2527 mol

Finally, we can calculate the mass of Li needed using its molar mass:

m(Li) = n(Li) x M(Li)

m(Li) = 0.2527 mol x 6.94 g/mol

m(Li) = 1.75 g

Therefore, 1.75 grams of lithium will combine with 75.0 grams of magnesium nitrate in this reaction.