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quester [9]
3 years ago
5

Can someone please show me how to do problem number 4? Please show work so I can try to understand it. Thanks!

Chemistry
1 answer:
larisa86 [58]3 years ago
5 0
With the given formula, we can calculate the amount of CO₂ using the balance equation but we first need the moles of CH₄

1) to find the moles of CH₄, we need to use the ideal gas formula (PV= nRT). if we solve for n, we solve for the moles of CH₄, and then we can convert to CO₂. Remember that the units put in this formula depending on the R value units. I remember 0.0821 which means pressure (P) has to be in atm, volume (V) in liters, the amount (n) in moles, and temperature (T) in kelvin.

PV= nRT

P= 1.00 atm
V= 32.0 Liters
n= ?
R= 0.0821 atm L/mol K
T= 25 C= 298 K

let plug the values into the formula.

(1.00 x 32.0 L)= n x 0.0821 x 298K

n= (1.00 x 32.0 L )/ (0.0821 x 298)= 1.31 moles CH₄

2) now let's convert the mole of CH₄ to moles to CO₂ using the balance equation

1.31 mol CH₄ (1 mol CO₂/ 1 mol CH₄)= 1.31 mol CO₂

3) Now let's convert from moles to grams using the molar mass of CO₂ (find the mass of each atom in the periodic table and add them)

molar mass CO₂= 12.00 + (2 x 16.0)= 44.0 g/mol

1.31 mol CO₂ ( 44.0 g/ 1 mol)= 57.6 g CO₂

Note: let me know if you any question.



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Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum
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Moles of copper are:

4.95cm^3\frac{8.95g}{1cm^3} \frac{1mol}{63.55g} = 0.697 moles

Moles of nitric acid are:

230mL\frac{1.42g}{mL} \frac{68g}{100g} \frac{1mol}{63.01g}=3.52moles

As 1 mol of Cu reacts with 4 moles of HNO₃:

0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.

Moles of NO₂ produced are:

0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>

Using PV = nRT

<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>

Thus, volume is:

V = nRT / P

V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm

V = 35.6L

<em>The volume of nitrogen oxide formed is 35.6L</em>

3 0
3 years ago
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