Answer:
See explanation.
Explanation:
Hello!
In this case, we can proceed as follows:
1. Here, the undergoing chemical reaction is:
[tex]C_2H_2+\frac{5}{2} O_2\rightarrow 2CO_2+H_2O[/tex]
Thus, the moles and mass of water turn out:
[tex]n_{H_2O}=20.0kgC_2H_2*\frac{1000gC_2H_2}{1kgC_2H_2} *\frac{1molC_2H_2}{26.04gC_2H_2} *\frac{1molH_2O}{1molC_2H_2}=768molH_2O\\\\m_{H_2O}=768molH_2O*\frac{18.02gH_2O}{1molH_2O}=13,840 gH_2O[/tex]
2. Here, the undergoing chemical reaction is:
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
So the required moles of HCl and the yielded of water are:
[tex]n_{CaCO_3}=2.6molHCl*\frac{1molCaCO_3}{2molHCl}=1.3molCaCO_3\\\\ n_{H_2O}=2.6molHCl*\frac{1molH_2O}{2molHCl}=1.3molH_2O[/tex]
3. Here, the undergoing chemical reaction is:
[tex]Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O[/tex]
Now, we apply each mole ratio obtain:
A.
[tex]n_{H_2SO_4}=2.6molAl_2O_3*\frac{3molH_2SO_4}{1molAl_2O_3} =7.8molH_2SO_4[/tex]
B.
[tex]n_{Al_2(SO_4)_3}=2.6molAl_2O_3*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3} =2.6molAl_2(SO_4)_3[/tex]
Best regards!
