What mass of carbon dioxide (co2) can be produced from 15.6 g of c6h14 and excess oxygen?

The mass of carbon dioxide produced is 46.7 grams.

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of hexane = 15.6

Molar mass of hexane = 88.2 g/mol

Putting values in equation 1, we get:

The chemical equation for the combustion of hexane follows:

As, oxygen gas is present in excess, it is considered as an excess reagent.

Thus, hexane is considered as a limiting reagent because it limits the formation of product

By Stoichiometry of the reaction:

2 moles of hexane produces 12 moles of carbon dioxide

So, 0.177 moles of hexane will produce = of carbon dioxide

Now, calculating the mass of carbon dioxide by using equation 1:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.062 moles

Putting values in equation 1, we get:

Hence, the mass of carbon dioxide produced is 46.7 grams.

explanation:

i think option no d is the correct answer

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