Question

1. balance the following reactions. indicate what type of reaction it is.

a. al + o₂ → al₂o₃
type of reaction___________
word equation:
b. caf₂ + h₂so₄ → caso₄ + hf
type of reaction____________________
word equation:
c. caco₃ → ca + co₃
type of reaction ____________________
word equation:

1. write a balanced chemical reaction for the following double displacement reactions:

d. iron (iii) phosphate reacts with sodium nitrate to form iron (iii) nitrate and sodium phosphate
e. magnesium acetate reacts with potassium hydroxide (you predict the products)
f. iron (i) chloride reacts with sodium hydroxide (you predict the products)

1. balance the following equation, and then answer the questions.

__ zn + hcl → zncl₂ + __ h₂

1. __ nacl + h₂so₄ → hcl + __ na₂so₄
2. calcium carbonate + lithium nitrate

build and balance the reaction

Explanation:

Step 1: Balance the equation for reaction 48

We start with the unbalanced equation: $Zn + HCl
ightarrow ZnCl_2 + H_2$

First, balance the chlorine atoms. On the right side, there are 2 Cl atoms in $ZnCl_2$, so we need 2 HCl on the left: $Zn + 2HCl
ightarrow ZnCl_2 + H_2$

Now check the other atoms. Zn is balanced (1 on each side), H: 2 on left (from 2HCl) and 2 on right (from $H_2$), Cl: 2 on left (from 2HCl) and 2 on right (from $ZnCl_2$). So the balanced equation is $1Zn + 2HCl
ightarrow 1ZnCl_2 + 1H_2$

Step 2: Balance the equation for reaction 49

Unbalanced equation: $NaCl + H_2SO_4
ightarrow HCl + Na_2SO_4$

Balance the Na atoms. On the right, there are 2 Na in $Na_2SO_4$, so we need 2 NaCl on the left: $2NaCl + H_2SO_4
ightarrow HCl + Na_2SO_4$

Now balance the Cl atoms. On the left, 2 Cl from 2NaCl, so we need 2 HCl on the right: $2NaCl + H_2SO_4
ightarrow 2HCl + Na_2SO_4$

Check other atoms: H: 2 on left (from $H_2SO_4$) and 2 on right (from 2HCl), S: 1 on each side, O: 4 on each side. So balanced equation is $2NaCl + 1H_2SO_4
ightarrow 2HCl + 1Na_2SO_4$

Step 3: Balance reaction 50 (calcium carbonate + lithium nitrate)

First, write the chemical formulas: $CaCO_3 + LiNO_3
ightarrow$

The products of a double displacement reaction would be $Ca(NO_3)_2 + Li_2CO_3$ (assuming a reaction occurs, though in reality, this reaction might not proceed as both products are soluble, but for balancing practice):

Unbalanced equation: $CaCO_3 + LiNO_3
ightarrow Ca(NO_3)_2 + Li_2CO_3$

Balance the Li atoms: On the right, 2 Li in $Li_2CO_3$, so we need 2 $LiNO_3$ on the left: $CaCO_3 + 2LiNO_3
ightarrow Ca(NO_3)_2 + Li_2CO_3$

Now check other atoms: Ca: 1 on each side, C: 1 on each side, O: 3 (from $CaCO_3$) + 6 (from 2$LiNO_3$) = 9; on the right, 6 (from $Ca(NO_3)_2$) + 3 (from $Li_2CO_3$) = 9. N: 2 on left (from 2$LiNO_3$) and 2 on right (from $Ca(NO_3)_2$). Li: 2 on left (from 2$LiNO_3$) and 2 on right (from $Li_2CO_3$). So balanced equation is $1CaCO_3 + 2LiNO_3
ightarrow 1Ca(NO_3)_2 + 1Li_2CO_3$

Answer:

s:

1. $1Zn + 2HCl

ightarrow 1ZnCl_2 + 1H_2$

1. $2NaCl + 1H_2SO_4

ightarrow 2HCl + 1Na_2SO_4$

1. $1CaCO_3 + 2LiNO_3

ightarrow 1Ca(NO_3)_2 + 1Li_2CO_3$ (Note: This reaction may not be favorable in reality as both products are soluble, but it is balanced for the given task)