Question

find the balanced coefficients of the following chemical reaction:
aqueous silver nitrate combines with
aqueous calcium bromide to produce solid
silver bromide and calcium nitrate
1,1,1,1
2,1,2,1
2,2,2,1
1,2,3,2

what is the balanced chemical equation for
the reaction of solid aluminum and
hydrochloric acid?
image of a document with chemical equations and options a, b, c, d
a
b
c
d

Explanation:

First Sub - Question (Balancing Silver Nitrate and Calcium Bromide Reaction)

Step 1: Write the unbalanced equation

The reaction is \(\ce{AgNO_3(aq) + CaBr_2(aq) -> AgBr(s) + Ca(NO_3)_2(aq)}\).

Step 2: Balance the bromide ions

On the left, we have 2 \(\ce{Br^-}\) from \(\ce{CaBr_2}\), and on the right, we have 1 \(\ce{Br^-}\) from \(\ce{AgBr}\). So we put a coefficient of 2 in front of \(\ce{AgBr}\), getting \(\ce{AgNO_3(aq) + CaBr_2(aq) -> 2AgBr(s) + Ca(NO_3)_2(aq)}\).

Step 3: Balance the silver ions

Now, on the right, we have 2 \(\ce{Ag^+}\) from \(2\ce{AgBr}\), so we put a coefficient of 2 in front of \(\ce{AgNO_3}\), resulting in \(\ce{2AgNO_3(aq) + CaBr_2(aq) -> 2AgBr(s) + Ca(NO_3)_2(aq)}\).

Step 4: Check the nitrate and calcium ions

The nitrate ions (\(\ce{NO_3^-}\)): on the left, 2 from \(2\ce{AgNO_3}\), on the right, 2 from \(\ce{Ca(NO_3)_2}\). Calcium ions: 1 on left (\(\ce{CaBr_2}\)) and 1 on right (\(\ce{Ca(NO_3)_2}\)). The equation is balanced with coefficients 2, 1, 2, 1.

Second Sub - Question (Aluminum and Hydrochloric Acid Reaction)

Step 1: Write the unbalanced equation

The reaction is \(\ce{Al(s) + HCl(aq) -> AlCl_3(aq) + H_2(g)}\).

Step 2: Balance the aluminum atoms

There is 1 Al on the left and 1 Al on the right initially, but to balance the chlorine and hydrogen, we will adjust. Let's start with Al. If we put 2 in front of \(\ce{Al}\) on the left, we need 2 in front of \(\ce{AlCl_3}\) on the right: \(2\ce{Al(s)} + \ce{HCl(aq)} -> 2\ce{AlCl_3(aq)} + \ce{H_2(g)}\).

Step 3: Balance the chlorine atoms

On the right, we have \(2\times3 = 6\) Cl atoms from \(2\ce{AlCl_3}\). So we put a 6 in front of \(\ce{HCl}\) on the left: \(2\ce{Al(s)} + 6\ce{HCl(aq)} -> 2\ce{AlCl_3(aq)} + \ce{H_2(g)}\).

Step 4: Balance the hydrogen atoms

On the left, we have 6 H from \(6\ce{HCl}\), on the right, we have 2 H from \(\ce{H_2}\). So we put a 3 in front of \(\ce{H_2}\) on the right: \(2\ce{Al(s)} + 6\ce{HCl(aq)} -> 2\ce{AlCl_3(aq)} + 3\ce{H_2(g)}\). This matches option c. Option a is unbalanced (Al and Cl are not balanced), option b has the wrong acid (it should be HCl, not \(\ce{HClO_3}\)), and option d is wrong because aluminum does react with acids.

Answer:

2, 1, 2, 1