Question

fill in the blank 4 points
1 li + 3 alcl₃ ⟶ 1 licl + 3
al

Explanation:

Step1: Identify the reaction type

This is a single - replacement reaction. The general form of a single - replacement reaction is \(A + BC
ightarrow AC + B\). Here, we have \(Li\) (lithium) reacting with \(AlCl_3\) (aluminum chloride).

Step2: Balance the equation

• Let's start by looking at the number of each atom on both sides.
• For \(Cl\) (chlorine) atoms: In \(AlCl_3\), there are 3 \(Cl\) atoms per formula unit. On the product side, in \(LiCl\), there is 1 \(Cl\) atom per formula unit. To balance the \(Cl\) atoms, we need 3 moles of \(LiCl\) for every 1 mole of \(AlCl_3\) (since \(3\times1 = 3\) for \(Cl\) in \(LiCl\) and \(1\times3=3\) for \(Cl\) in \(AlCl_3\)).
• For \(Al\) (aluminum) atoms: In \(AlCl_3\), there is 1 \(Al\) atom per formula unit. On the product side, we should have 1 \(Al\) atom. So the coefficient of \(Al\) is 1.
• For \(Li\) (lithium) atoms: Since we have 3 moles of \(LiCl\) (from the \(Cl\) balance), and each \(LiCl\) has 1 \(Li\) atom, we need 3 moles of \(Li\) on the reactant side.

The balanced chemical equation is \(3Li+AlCl_3
ightarrow3LiCl + Al\). Wait, but in the given blanks, let's check the existing numbers. The first blank (for \(Li\)): if we follow the correct balancing, the coefficient of \(Li\) should be 3, the coefficient of \(AlCl_3\) is 1, the coefficient of \(LiCl\) is 3, and the coefficient of \(Al\) is 1. But in the given problem, the blanks are filled with 1 (for \(Li\)), 3 (for \(AlCl_3\)), 1 (for \(LiCl\)) and 3 (for the last blank, which should be \(Al\)). Wait, maybe there is a mis - balancing in the problem's initial filling. But if we correct it, the balanced equation is \(3Li+AlCl_3 = 3LiCl+Al\). But according to the blanks provided, the last blank should be filled with \(Al\) (aluminum) with a coefficient of 1? Wait, no, let's re - examine.

Wait, the reaction is a single - displacement reaction where \(Li\) displaces \(Al\) from \(AlCl_3\). The correct balancing:

• Reactants: \(Li\) and \(AlCl_3\)
• Products: \(LiCl\) and \(Al\)

Let's balance the atoms:

• \(Al\): 1 on left (\(AlCl_3\)), so 1 on right (\(Al\))
• \(Cl\): 3 on left (\(AlCl_3\)), so we need 3 \(Cl\) on right. Since each \(LiCl\) has 1 \(Cl\), we need 3 \(LiCl\)
• \(Li\): 3 on right (from 3 \(LiCl\)), so we need 3 \(Li\) on left.

So the balanced equation is \(3Li + AlCl_3=3LiCl + Al\). But in the given problem, the first blank (for \(Li\)) has 1, the second (for \(AlCl_3\)) has 3, the third (for \(LiCl\)) has 1, and the fourth blank (the last one) should be \(Al\) with a coefficient of 1? Wait, maybe the problem has a typo, but if we go by the blanks, the last blank is for \(Al\), and the coefficient should be 1. But according to the correct balancing, the coefficients are 3 (Li), 1 (AlCl₃), 3 (LiCl), 1 (Al).

Answer:

The last blank should be filled with \(Al\) (aluminum) with a coefficient of 1. So the completed equation is \(3Li + AlCl_3 = 3LiCl+Al\) (but if we follow the given initial numbers in the blanks, the last blank is for \(Al\) and the coefficient is 1, and the coefficient of \(Li\) should be 3, coefficient of \(AlCl_3\) is 1, coefficient of \(LiCl\) is 3). However, based on the blanks provided in the problem (with 1 for Li, 3 for \(AlCl_3\), 1 for \(LiCl\)), the last blank (the product) should be \(Al\) with a coefficient of 1. Wait, no, let's re - calculate.

Wait, the number of \(Cl\) atoms: on the left, \(3\times3 = 9\) (since \(AlCl_3\) has coefficient 3), on the right, \(LiCl\) has coefficient 1, so \(Cl\) atoms on right is 1. That's not balanced. So there is a mistake in the initial numbers in the blanks. But if we assume that the problem is trying to balance with the given blanks, and the last product is \(Al\), the coefficient of \(Al\) should be 1 (since \(AlCl_3\) has 1 \(Al\) atom). So the last blank is \(Al\) with a coefficient of 1. But according to the correct balancing, the coefficients should be adjusted. But based on the problem's layout, the last blank is for the element \(Al\), so the answer for the last blank is \(Al\) (aluminum) with a coefficient of 1 (but the initial coefficient for \(Li\) should be 3, \(AlCl_3\) should be 1, \(LiCl\) should be 3, and \(Al\) should be 1).