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kondaur [170]
3 years ago
12

E is solution of HCI of unk concentration f contain 4.8g of NaOH in 25cm point of salution​

Chemistry
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

18

Explanation:

no explanation sorry

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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)}

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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)

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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)

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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

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