q.5. balance the following chemical equations.\na) __alcl₃ + __h₂o → __al(oh)₃ + __hcl\nb) __c₃h₈ + __o₂ →…
q.5. balance the following chemical equations.\na) __alcl₃ + __h₂o → __al(oh)₃ + __hcl\nb) __c₃h₈ + __o₂ → __co₂ + __h₂o\nc) __al + __feo → __al₂o₃ + __fe\nd) __s₈ + __o₂ → __so₃\ne) __c₇h₆o₂ + __o₂ → __co₂ + __h₂o\nf) __baf₂ + __k₃po₄ → __ba₃(po₄)₂ + __kf\ng) __fecl₃ + __naoh → __fe(oh)₃ + __nacl\nh) __k₃po₄ + __hcl → __kcl + __h₃po₄\ni) __cabr₂ + __na₃p → __ca₃p₂ + __nabr\nj) __b₂br₆ + __hno₃ → __b(no₃)₃ + __hbr
Answer
Explanation:
Step1: Balance equation h
For the reaction $K_3PO_4 + HCl\rightarrow KCl + H_3PO_4$, balance potassium first. There are 3 potassium atoms in $K_3PO_4$, so we need 3 moles of $KCl$. Then, to balance chlorine, since we have 3 moles of $KCl$, we need 3 moles of $HCl$. The balanced equation is $1K_3PO_4+3HCl\rightarrow3KCl + 1H_3PO_4$.
Step2: Balance equation j
For the reaction $B_2Br_6+HNO_3\rightarrow B(NO_3)_3 + HBr$, balance boron first. There are 2 boron atoms in $B_2Br_6$, so we need 2 moles of $B(NO_3)_3$. Then, for the nitrate - ions, since we have 2 moles of $B(NO_3)_3$, we need 6 moles of $HNO_3$. Finally, for bromine, since we have 6 moles of bromine in $B_2Br_6$, we need 6 moles of $HBr$. The balanced equation is $1B_2Br_6 + 6HNO_3\rightarrow2B(NO_3)_3+6HBr$.
Answer:
a) $1AlCl_3 + 3H_2O\rightarrow1Al(OH)_3+3HCl$ b) $1C_3H_8 + 5O_2\rightarrow3CO_2 + 4H_2O$ c) $2Al+3FeO\rightarrow1Al_2O_3 + 3Fe$ d) $1S_8+12O_2\rightarrow8SO_3$ e) $2C_7H_6O_2+15O_2\rightarrow14CO_2+6H_2O$ f) $3BaF_2+2K_3PO_4\rightarrow1Ba_3(PO_4)_2 + 6KF$ g) $1FeCl_3+3NaOH\rightarrow1Fe(OH)_3+3NaCl$ h) $1K_3PO_4+3HCl\rightarrow3KCl + 1H_3PO_4$ i) $3CaBr_2+2Na_3P\rightarrow1Ca_3P_2+6NaBr$ j) $1B_2Br_6 + 6HNO_3\rightarrow2B(NO_3)_3+6HBr$