Question

1. 5.00 g of each reactant are used. determine the theoretical yield of nh₃.

note: this problem is both limiting reactant and sequential reactions.
(1.56 g nh₃ vs 1.46 g nh₃ vs 14.2 g nh₃ vs 6.08 g nh₃; answer = 1.46 g nh₃)

Explanation:

Step1: Calculate moles of \(H_2O\) from first reaction

Molar mass of \(H_2O\) is \(M_{H_2O}=18.02\ g/mol\). Moles of \(H_2O\) from \(Al_2(CO_3)_3\) reaction: \(n_{H_2O}=\frac{5.00\ g}{18.02\ g/mol}\approx0.277\ mol\)

Step2: Calculate moles of \(H_2\) from \(CH_4\) reaction

From \(2H_2O + CH_4
ightarrow CO_2+4H_2\), mole ratio \(H_2O:H_2 = 2:4 = 1:2\). Moles of \(H_2\) from \(H_2O\): \(n_{H_2}=0.277\ mol\times2 = 0.554\ mol\)

Step3: Calculate moles of \(H_2\) from \(N_2\) reaction

Molar mass of \(H_2\) is \(M_{H_2}=2.02\ g/mol\). Moles of \(H_2\) available: \(n_{H_2}=\frac{5.00\ g}{2.02\ g/mol}\approx2.475\ mol\) (but we take the smaller amount from previous step as limiting)

Step4: Calculate moles of \(NH_3\)

From \(3H_2 + N_2
ightarrow2NH_3\), mole ratio \(H_2:NH_3=3:2\). Moles of \(NH_3\): \(n_{NH_3}=0.554\ mol\times\frac{2}{3}\approx0.369\ mol\)

Step5: Calculate mass of \(NH_3\)

Molar mass of \(NH_3\) is \(M_{NH_3}=17.03\ g/mol\). Mass of \(NH_3\): \(m_{NH_3}=0.369\ mol\times17.03\ g/mol\approx6.28\ g\) (Wait, no, re - check. Wait, actually, we need to consider all steps properly.

Let's start over:

Step1: First reaction \(6HCl + Al_2(CO_3)_3

ightarrow2AlCl_3 + 3H_2O+3CO_2\)
Molar mass of \(Al_2(CO_3)_3\): \(M = 2\times26.98+(3\times12.01)+(9\times16.00)=233.99\ g/mol\). Moles of \(Al_2(CO_3)_3\): \(n=\frac{5.00\ g}{233.99\ g/mol}\approx0.0214\ mol\). Moles of \(H_2O\) produced: \(n_{H_2O}=0.0214\ mol\times3 = 0.0642\ mol\)

Step2: Second reaction \(2H_2O+CH_4

ightarrow CO_2 + 4H_2\)
Molar mass of \(CH_4\): \(M = 16.04\ g/mol\). Moles of \(CH_4\): \(n=\frac{5.00\ g}{16.04\ g/mol}\approx0.312\ mol\). From \(H_2O\): moles of \(H_2\) produced \(n_{H_2}=0.0642\ mol\times2=0.128\ mol\) (since \(H_2O\) is limiting here as \(0.0642\ mol\) \(H_2O\) vs \(0.312\ mol\) \(CH_4\))

Step3: Third reaction \(3H_2+N_2

ightarrow2NH_3\)
Molar mass of \(N_2\): \(M = 28.02\ g/mol\). Moles of \(N_2\): \(n=\frac{5.00\ g}{28.02\ g/mol}\approx0.178\ mol\). Moles of \(H_2\) from step 2: \(n_{H_2}=0.128\ mol\). From \(H_2\) - \(N_2\) reaction, mole ratio \(H_2:NH_3 = 3:2\). Moles of \(NH_3\): \(n_{NH_3}=0.128\ mol\times\frac{2}{3}\approx0.0853\ mol\). Mass of \(NH_3\): \(m = 0.0853\ mol\times17.03\ g/mol\approx1.46\ g\)

Answer:

\(1.46\ g\) \(NH_3\)