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3 years ago

how many moles of methane are produced when 25.1 moles of carbon dioxide gas react with excess hydrogen

Chemistry

2 answers:

Tanya [424]3 years ago

7 0

The moles of methane that are produced when 25.1 moles of CO2 gas react with excess H2 is 25.1 moles

calculation

write the equation for the reaction

that is CO2 +4 H2 → CH4 +2H2O

from the equation above the mole ratio between CO2 to CH4 is 1:1 therefore the moles of CH4 is also= 25.1 moles

Dafna11 [192]3 years ago

6 0

The chemical equation for this reaction is this:
CO2 + 4H2 = CH4 + 2H2O
from the equation, the mole ration of CO2 to CH4 is ratio 1:1
and CH4 = 25.1
The mole ratio of H2 to H2O is 4:2
therefore, mole of H2 = 19.1* 4/2 = 38.2
The moles of methane produced is 38.2 moles

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Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 $\frac{mol}{L}$

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL

$Volume of solution = 725 mL * \frac{1L }{1000 mL}$

Volume of solution = 0.725 L

Following steps can be done to find mass of :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

$Na_3PO_4 3 Na^+ + PO_4^3^-$

Step 2: Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula which is : $Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }$

Plugging value of Molarity and volume

$1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}$

Multiplying both side by 0.725 L

$1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L$

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion = 3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

$Mole of Na_3PO_4= Mole of Na^+ * Mole ratio$

$Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }$

Mole of Na₃PO₄ = 0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :

$Mass (g) = mole (mol) * molar mass \frac{g}{mol}$

$Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}$

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8 0

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Answer:

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Given data:

Volume of gas = 5.0 L

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