C2h6 (g) + o2 (g) → co2 (g) + h2o (g) (not balanced) determine the number of liters of co2 formed at stp when 240.0 grams of c2h6 is burned in excess oxygen gas?

13.44dm^3

Explanation:

To calculate this we first need to know the number of mole produced. We will first need to balance the equation to know the theoretical mole ratio.

C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (g)

From the balanced equation, we can deduce that one mole of ethane yielded 2 moles of carbon iv oxide. We use this information to calculate the actual number of moles yielded.

24g were reacted. Now to know the number of moles reacted, we simply divide the mass by the molar mass. The molar mass of ethane is 2(12) +6(1)= 40g/mol

The number of moles is thus 24/40 = 0.6 moles

Like we said earlier, one mole yielded 2 moles of carbon iv oxide, hence, 0.6 moles will yield 0.6 * 2 = 0.12 moles of carbon iv oxide.

Now, at stp, one mole of a gas occupies a volume of 22.4dm^3 thus, 0.6 mole will occupy 0.6 * 22.4 = 13.44dm^3

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

explanation: