Question

CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST

Answer 1

Explanation:

2. $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

First, we need to find the number of moles of $CO_2$ at 300K and 1.5 atm using the ideal gas law:

$n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}$

$=2.0\:\text{mol}\:CO_2$

Now use the molar ratios to find the number of moles of ethane to produce this much $CO_2$.

$2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)$

$=1.0\:\text{mol}\:C_2H_6$

Finally, convert this amount to grams using its molar mass:

$1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)$

$=30.1\:g\:C_2H_6$

3. $3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2$

Convert 75 g Zn into moles:

$75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn$

Then use the molar ratios to find the amount of H2 produced.

$1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2$

Now use the ideal gas law $PV=nRT$ to find the volume of H2 produced at 23°C and 4 atm:

$V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}$

$=8.9\:\text L\:H_2$