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balance the following equations 11. ​cu + ​agno3 → ​cu(no3)2 + ​ag 12. …

Question

balance the following equations

  1. ​cu + ​agno3 → ​cu(no3)2 + ​ag
  2. ​(nh4)3po4 + ​ba(oh)2 → ​ba3(po4)2 + ​nh4oh
  3. ​nh3 + ​o2 → ​n2 + ​h2o

provide the molar masses for the following

  1. k ​
  2. n3cl ​
  3. pbo2 ​
  4. mg(no3)2 ​

use the molar mass formula for the following questions

  1. determine the mass of 4.2mol of calcium ​
  2. determine the mass of 6.27 x 10-3mol (.00627) of ba(no2)2 ​
  3. how many mols of lithium are there in 24.29g ​
  4. how many mols of naoh are there in 390g ​

Explanation:

Step1: Balance the equation \(Cu + AgNO_3

ightarrow Cu(NO_3)_2+Ag\)

  • For the nitrate ion \((NO_3^-)\): In \(Cu(NO_3)_2\) there are 2 nitrate ions. So, put a 2 in front of \(AgNO_3\).
  • Now, for Ag: Since there are 2 \(Ag\) atoms from \(2AgNO_3\), put a 2 in front of \(Ag\).
  • The balanced equation is \(Cu + 2AgNO_3

ightarrow Cu(NO_3)_2+2Ag\)

Step2: Balance the equation \((NH_4)_3PO_4 + Ba(OH)_2

ightarrow Ba_3(PO_4)_2+NH_4OH\)

  • For \(Ba\): In \(Ba_3(PO_4)_2\) there are 3 \(Ba\) atoms. So, put a 3 in front of \(Ba(OH)_2\).
  • For \(PO_4^{3 - }\): In \(Ba_3(PO_4)_2\) there are 2 phosphate ions. So, put a 2 in front of \((NH_4)_3PO_4\).
  • Now, for \(NH_4^+\) and \(OH^-\): There are 6 \(NH_4^+\) and 6 \(OH^-\) from \(2(NH_4)_3PO_4\) and \(3Ba(OH)_2\). So, put a 6 in front of \(NH_4OH\).
  • The balanced equation is \(2(NH_4)_3PO_4 + 3Ba(OH)_2

ightarrow Ba_3(PO_4)_2+6NH_4OH\)

Step3: Balance the equation \(NH_3+O_2

ightarrow N_2 + H_2O\)

  • For \(N\): There are 2 \(N\) atoms in \(N_2\). So, put a 2 in front of \(NH_3\).
  • For \(H\): There are 6 \(H\) atoms from \(2NH_3\). So, put a 3 in front of \(H_2O\).
  • For \(O\): There are 3 \(O\) atoms in \(3H_2O\). So, put \(\frac{3}{2}\) in front of \(O_2\). But we usually use whole - number coefficients. Multiply all coefficients by 2.
  • The balanced equation is \(4NH_3+3O_2

ightarrow 2N_2 + 6H_2O\)

Step4: Calculate molar mass of \(K\)

  • The atomic mass of \(K\) (potassium) is approximately \(39.10\space g/mol\)

Step5: Calculate molar mass of \(N_3Cl\)

  • The atomic mass of \(N\) is approximately \(14.01\space g/mol\) and of \(Cl\) is approximately \(35.45\space g/mol\).
  • \(M = 3\times14.01+35.45=42.03 + 35.45=77.48\space g/mol\)

Step6: Calculate molar mass of \(PbO_2\)

  • The atomic mass of \(Pb\) is approximately \(207.2\space g/mol\) and of \(O\) is approximately \(16.00\space g/mol\).
  • \(M=207.2 + 2\times16.00=207.2+32.00 = 239.2\space g/mol\)

Step7: Calculate molar mass of \(Mg(NO_3)_2\)

  • The atomic mass of \(Mg\) is approximately \(24.31\space g/mol\), of \(N\) is approximately \(14.01\space g/mol\), and of \(O\) is approximately \(16.00\space g/mol\).
  • \(M = 24.31+2\times(14.01 + 3\times16.00)=24.31+2\times(14.01+48.00)=24.31+2\times62.01=24.31 + 124.02=148.33\space g/mol\)

Step8: Determine the mass of \(4.2\space mol\) of Calcium

  • The molar mass of \(Ca\) is approximately \(40.08\space g/mol\).
  • Using the formula \(m=n\times M\), where \(n = 4.2\space mol\) and \(M = 40.08\space g/mol\)
  • \(m=4.2\times40.08 = 168.34\space g\)

Step9: Determine the mass of \(6.27\times10^{-3}\space mol\) of \(Ba(NO_2)_2\)

  • The molar mass of \(Ba(NO_2)_2\):
  • Atomic mass of \(Ba = 137.33\space g/mol\), \(N = 14.01\space g/mol\), \(O=16.00\space g/mol\)
  • \(M=137.33+2\times(14.01 + 2\times16.00)=137.33+2\times(14.01 + 32.00)=137.33+2\times46.01=137.33 + 92.02=229.35\space g/mol\)
  • Using \(m=n\times M\), where \(n = 6.27\times10^{-3}\space mol\) and \(M = 229.35\space g/mol\)
  • \(m=6.27\times10^{-3}\times229.35\approx1.44\space g\)

Step10: Calculate the number of moles of Lithium in \(24.29\space g\)

  • The molar mass of \(Li\) is approximately \(6.94\space g/mol\)
  • Using \(n=\frac{m}{M}\), where \(m = 24.29\space g\) and \(M = 6.94\space g/mol\)
  • \(n=\frac{24.29}{6.94}\approx3.50\space mol\)

Step11: Calculate the number of moles of \(NaOH\) in \(390\space g\)

  • The molar mass of \(NaOH\): \(M = 22.99+16.00 + 1.01=40.00\space g/mol\)
  • Using \(n=\frac{m}{M}\), where \(m = 390\space g\) and \(M = 40.00\space g/mol\)
  • \(n=\frac{390}{40.00}=9.75\space mol\)

Answer:

  1. \(1Cu + 2AgNO_3

ightarrow 1Cu(NO_3)_2+2Ag\)

  1. \(2(NH_4)_3PO_4 + 3Ba(OH)_2

ightarrow 1Ba_3(PO_4)_2+6NH_4OH\)

  1. \(4NH_3+3O_2

ightarrow 2N_2 + 6H_2O\)

  1. \(39.10\space g/mol\)
  2. \(77.48\space g/mol\)
  3. \(239.2\space g/mol\)
  4. \(148.33\space g/mol\)
  5. \(168.34\space g\)
  6. \(1.44\space g\)
  7. \(3.50\space mol\)
  8. \(9.75\space mol\)