practice: balancing and classifying\nshow all work for balancing on a separate sheet of paper. place…

practice: balancing and classifying\nshow all work for balancing on a separate sheet of paper. place coefficient final answers in the blanks provided. you can leave coefficients of 1 blank. in the space to the right of each equation, classify the type of reaction\n1. ___ albr₃ + ___ k → ___ kbr + ___ al\n2. ___ feo + ___ pdf₂ → ___ fef₂ + ___ pdo\n3. ___ p₄ + ___ br₂ → ___ pbr₃\n4. ___ licl + ___ br₂ → ___ libr + ___ cl₂\n5. ___ pbbr₂ + ___ hcl → ___ hbr + ___ pbcl₂\n6. ___ cobr₃ + ___ caso₄ → ___ cabr₂ + ___ co₂(so₄)₃\n7. ___ na₃p + ___ caf₂ → ___ naf + ___ ca₃p₂\n8. ___ mn + ___ hi → ___ h₂ + ___ mni₃\n9. ___ li₃po₄ + ___ nabr → ___ na₃po₄ + ___ libr\n10. ___ caf₂ + ___ li₂so₄ → ___ caso₄ + ___ lif\n11. ___ hbr + ___ mg(oh)₂ → ___ mgbr₂ + ___ h₂o\n12. ___ lino₃ + ___ cabr₂ → ___ ca(no₃)₂ + ___ libr\n13. ___ agno₃ + ___ li → ___ lino₃ + ___ ag\n14. ___ si(oh)₄ + ___ nabr → ___ sibr₄ + ___ naoh\n15. ___ nacn + ___ cuco₃ → ___ na₂co₃ + ___ cu(cn)₂

practice: balancing and classifying\nshow all work for balancing on a separate sheet of paper. place coefficient final answers in the blanks provided. you can leave coefficients of 1 blank. in the space to the right of each equation, classify the type of reaction\n1. ___ albr₃ + ___ k → ___ kbr + ___ al\n2. ___ feo + ___ pdf₂ → ___ fef₂ + ___ pdo\n3. ___ p₄ + ___ br₂ → ___ pbr₃\n4. ___ licl + ___ br₂ → ___ libr + ___ cl₂\n5. ___ pbbr₂ + ___ hcl → ___ hbr + ___ pbcl₂\n6. ___ cobr₃ + ___ caso₄ → ___ cabr₂ + ___ co₂(so₄)₃\n7. ___ na₃p + ___ caf₂ → ___ naf + ___ ca₃p₂\n8. ___ mn + ___ hi → ___ h₂ + ___ mni₃\n9. ___ li₃po₄ + ___ nabr → ___ na₃po₄ + ___ libr\n10. ___ caf₂ + ___ li₂so₄ → ___ caso₄ + ___ lif\n11. ___ hbr + ___ mg(oh)₂ → ___ mgbr₂ + ___ h₂o\n12. ___ lino₃ + ___ cabr₂ → ___ ca(no₃)₂ + ___ libr\n13. ___ agno₃ + ___ li → ___ lino₃ + ___ ag\n14. ___ si(oh)₄ + ___ nabr → ___ sibr₄ + ___ naoh\n15. ___ nacn + ___ cuco₃ → ___ na₂co₃ + ___ cu(cn)₂

Answer

Explanation:

Step1: Balance equation 1

$2AlBr_3 + 6K \rightarrow 6KBr+2Al$, single - replacement reaction

Step2: Balance equation 2

$FeO + PdF_2 \rightarrow FeF_2+PdO$, double - replacement reaction

Step3: Balance equation 3

$P_4 + 6Br_2 \rightarrow 4PBr_3$, synthesis reaction

Step4: Balance equation 4

$2LiCl + Br_2 \rightarrow 2LiBr + Cl_2$, single - replacement reaction

Step5: Balance equation 5

$PbBr_2 + 2HCl \rightarrow 2HBr+PbCl_2$, double - replacement reaction

Step6: Balance equation 6

$2CoBr_3 + 3CaSO_4 \rightarrow 3CaBr_2+Co_2(SO_4)_3$, double - replacement reaction

Step7: Balance equation 7

$2Na_3P+ 3CaF_2 \rightarrow 6NaF + Ca_3P_2$, double - replacement reaction

Step8: Balance equation 8

$2Mn + 6HI \rightarrow 3H_2+2MnI_3$, single - replacement reaction

Step9: Balance equation 9

$Li_3PO_4+3NaBr \rightarrow Na_3PO_4 + 3LiBr$, double - replacement reaction

Step10: Balance equation 10

$CaF_2+Li_2SO_4 \rightarrow CaSO_4 + 2LiF$, double - replacement reaction

Step11: Balance equation 11

$2HBr+Mg(OH)_2 \rightarrow MgBr_2 + 2H_2O$, double - replacement reaction

Step12: Balance equation 12

$2LiNO_3+CaBr_2 \rightarrow Ca(NO_3)_2+2LiBr$, double - replacement reaction

Step13: Balance equation 13

$AgNO_3+Li \rightarrow LiNO_3+Ag$, single - replacement reaction

Step14: Balance equation 14

$Si(OH)_4 + 4NaBr \rightarrow SiBr_4+4NaOH$, double - replacement reaction

Step15: Balance equation 15

$2NaCN+CuCO_3 \rightarrow Na_2CO_3+Cu(CN)_2$, double - replacement reaction

Answer:

  1. 2, 6, 6, 2; single - replacement
  2. 1, 1, 1, 1; double - replacement
  3. 1, 6, 4; synthesis
  4. 2, 1, 2, 1; single - replacement
  5. 1, 2, 2, 1; double - replacement
  6. 2, 3, 3, 1; double - replacement
  7. 2, 3, 6, 1; double - replacement
  8. 2, 6, 3, 2; single - replacement
  9. 1, 3, 1, 3; double - replacement
  10. 1, 1, 1, 2; double - replacement
  11. 2, 1, 1, 2; double - replacement
  12. 2, 1, 1, 2; double - replacement
  13. 1, 1, 1, 1; single - replacement
  14. 1, 4, 1, 4; double - replacement
  15. 2, 1, 1, 1; double - replacement