Explanation:
2. [tex]2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O[/tex]
First, we need to find the number of moles of [tex]CO_2[/tex] at 300K and 1.5 atm using the ideal gas law:
[tex]n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}[/tex]
[tex]=2.0\:\text{mol}\:CO_2[/tex]
Now use the molar ratios to find the number of moles of ethane to produce this much [tex]CO_2[/tex].
[tex]2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)[/tex]
[tex]=1.0\:\text{mol}\:C_2H_6[/tex]
Finally, convert this amount to grams using its molar mass:
[tex]1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)[/tex]
[tex]=30.1\:g\:C_2H_6[/tex]
3. [tex]3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2[/tex]
Convert 75 g Zn into moles:
[tex]75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn[/tex]
Then use the molar ratios to find the amount of H2 produced.
[tex]1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2[/tex]
Now use the ideal gas law [tex]PV=nRT[/tex] to find the volume of H2 produced at 23°C and 4 atm:
[tex]V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}[/tex]
[tex]=8.9\:\text L\:H_2[/tex]
