Answer:
1) Br2 is the limiting reactant.
Mass NaBr produced = 64.4 grams
2) CuSO4 is the limiting reactant
Mass Cu = 19.89 grams
Mass ZnSO4 = 50.54 grams
3) NH4Cl is the limiting reactant
Mass NaCl = 54.6 grams
Mass NH3 =15.9 grams
Mass H2O =16.8 grams
4) Fe2O3 is the limiting reactant
Mass Fe = 35.0 grams
Mass CO2 = 41.3 grams
Explanation:
1) Na+br2 ------------->Nabr
Step 1: Data given
Mass Na = 50.0 grams
Mass Br2 = 50.0 grams
Molar mass Na = 22.99 g/mol
Molar mass Br2 = 159.81 g/mol
Step 2: The balanced equation
2Na + Br2 → 2NaBr
Step 3: Calculate moles
Moles = mass / molar mass
Moles Na = 50.0 grams / 22.99 g/mol = 2.17 moles
Moles Br2 = 50.0 grams / 159.81 g/mol = 0.313 moles
Step 4: Calculate limiting reactant
Br2 is the limiting reactant. It will completely be consumed (0.313 moles).
Na is in excess. There will react 2*0.313 = 0.626 moles
There will remain 2.17 - 0.626 = 1.544 moles
Step 5: Calculate moles NaBr
For 1 mol Br2 we'll have 2 moles NaBr
For 0.313 moles we'll have 0.626 moles NaBr
Step 6: Calculate mass NaBr
Mass NaBr = 0.626 moles * 102.89 g/mol
Mass NaBr = 64.4 grams
2) Zn+cuso4 -------------->Znso4+Cu
Step 1: Data given
Mass Zn = 50.0 grams
Mass CuSO4 = 50.0 grams
Molar mass Zn = 65.38 g/mol
Molar mass CuSO4 = 159.61 g/mol
Step 2: The balanced equation
Zn + CuSO4 → Cu + ZnSO4
Step 3: Calculate moles
Moles = mass / molar mass
Moles Zn = 50.0 grams / 65.38 g/mol = 0.765 moles
Moles CuSO4 = 50.0 grams / 159.61 g/mol = 0.313 moles
Step 4: Calculate limiting reactant
CuSO4 is the limiting reactant. It will completely be consumed (0.313 moles).
Zn is in excess. There will react 0.313 moles
There will remain 0.765 - 0.313 = 0.452 moles
Step 5: Calculate moles products
For 1 mol Zn we need 1 mol CuSO4 to produce 1 mol Cu and 1 mol ZnSO4
For 0.313 moles CuSO4 we'll have 0.313 moles Cu and 0.313 moles ZnSO4
Step 6: Calculate mass products
Mass Cu = 0.313 moles * 63.546 g/mol = 19.89 grams
Mass ZnSO4 = 0.313 moles * 161.47 g/mol = 50.54 grams
3) NH4cl+NaOH -------------->NH3+H2O+NaCl
Step 1: Data given
Mass NH4Cl = 50.0 grams
Mass NaOH = 50.0 grams
Molar mass NH4Cl = 53.49 g/mol
Molar mass NaOH = 40.0 g/mol
Step 2: The balanced equation
NH4Cl + NaOH → NaCl + NH3 + H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles NH4Cl = 50.0 grams / 53.49 g/mol = 0.935 moles
Moles NaOH = 50.0 grams / 40.0 g/mol = 1.25 moles
Step 4: Calculate limiting reactant
NH4Cl is the limiting reactant. It will completely be consumed (0.935 moles).
NaOH is in excess. There will react 0.935 moles
There will remain 1.25 - 0.935 = 0.315 moles
Step 5: Calculate moles products
For 1 mol NH4Cl we need 1 mol NaOH to produce 1 mol NaCl, 1 mol NH3 and 1 mol H2O
For 0.935 moles NH4Cl we'll have 0.935 moles NaCl, 0.935 moles NH3 and 0.935 moles H2O
Step 6: Calculate mass products
Mass NaCl = 0.935 moles * 58.44 g/mol = 54.6 grams
Mass NH3 = 0.935 moles * 17.03 g/mol = 15.9 grams
Mass H2O = 0.935 moles * 18.02 g/mol = 16.8 grams
4) Fe2O3+CO ------------>Fe+CO2
Step 1: Data given
Mass Fe2O3 = 50.0 grams
Mass CO = 50.0 grams
Molar mass Fe2O3 = 159.69 g/mol
Molar mass CO = 28.01 g/mol
Step 2: The balanced equation
Fe2O3 + 3CO → 2Fe + 3CO2
Step 3: Calculate moles
Moles = mass / molar mass
Moles Fe2O3 = 50.0 grams / 159.69 g/mol = 0.313 moles
Moles CO = 50.0 grams / 28.01 g/mol = 1.785 moles
Step 4: Calculate limiting reactant
Fe2O3 is the limiting reactant. It will completely be consumed (0.313 moles).
CO is in excess. There will react 3* 0.313 = 0.939 moles
There will remain 1.785 - 0.939 = 0.846 moles
Step 5: Calculate moles products
For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe, 3 moles CO2
For 0.313 moles Fe2O3 we'll have 0.626 moles Fe and 0.939 moles CO2
Step 6: Calculate mass products
Mass Fe = 0.626 moles * 55.845 g/mol = 35.0 grams
Mass CO2 = 0.939 moles * 44.01 g/mol = 41.3 grams
