science grade 10\nassignment 2 - chemical reactions\nname: \nappl. /22\n1. balance each equation and…

science grade 10\nassignment 2 - chemical reactions\nname: \nappl. /22\n1. balance each equation and indicate the type of reaction (for e.g.: synthesis, decomposition, single displacement, double displacement)\na - 10 marks\na) pb(no₃)₂ + al₂(so₄)₃ → pbso₄ + al(no₃)₃\ntype: \nb) br₂ + o₂ → br₂o₃\ntype: \nc) cr₂o₃ + c → cr + co₂\ntype: \nd) na + h₂so₄ → h₂ + na₂so₄\ntype:
Answer
Explanation:
Step1: Balance equation a
For $Pb(NO_3)_2+Al_2(SO_4)_3\rightarrow PbSO_4 + Al(NO_3)_3$, balance $Pb$ first, then $Al$, $SO_4^{2 -}$ and $NO_3^{-}$. The balanced equation is $3Pb(NO_3)_2+Al_2(SO_4)_3 = 3PbSO_4+2Al(NO_3)_3$. This is a double - displacement reaction as the cations ($Pb^{2+}$ and $Al^{3+}$) exchange anions ($NO_3^{-}$ and $SO_4^{2 -}$).
Step2: Balance equation b
For $Br_2+O_2\rightarrow Br_2O_3$, balance the oxygen atoms. Multiply $O_2$ by 3 and $Br_2O_3$ by 2, then $Br_2$ by 2. The balanced equation is $2Br_2 + 3O_2=2Br_2O_3$. This is a synthesis reaction as two elements combine to form a single compound.
Step3: Balance equation c
For $Cr_2O_3 + C\rightarrow Cr+CO_2$, balance $Cr$ first, then $O$ and $C$. Multiply $Cr_2O_3$ by 2, $CO_2$ by 3, $C$ by 3 and $Cr$ by 4. The balanced equation is $2Cr_2O_3+3C = 4Cr + 3CO_2$. This is a single - displacement reaction where carbon displaces chromium from its oxide.
Step4: Balance equation d
For $Na+H_2SO_4\rightarrow H_2+Na_2SO_4$, balance $Na$ first. Multiply $Na$ by 2. The balanced equation is $2Na+H_2SO_4=H_2 + Na_2SO_4$. This is a single - displacement reaction where sodium displaces hydrogen from sulfuric acid.
Answer:
a) Balanced equation: $3Pb(NO_3)_2+Al_2(SO_4)_3 = 3PbSO_4+2Al(NO_3)_3$, type: double - displacement b) Balanced equation: $2Br_2 + 3O_2=2Br_2O_3$, type: synthesis c) Balanced equation: $2Cr_2O_3+3C = 4Cr + 3CO_2$, type: single - displacement d) Balanced equation: $2Na+H_2SO_4=H_2 + Na_2SO_4$, type: single - displacement