Be sure to answer all parts. industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. the products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane

Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

M(C₃H₈) = 44.1 g/mol; molar mass of propane.

M(H₂) = 2 g/mol; molar mass of hydrogen.

From balanced chemical reaction: n(C₃H₈) : n(H₂) = 1 : 7.

7m(C₃H₈) : M(C₃H₈) = m(H₂) : M(H₂).

7·8310 kg : 44.1 g/mol = m(H₂) : 2 g/mol.

m(H₂) = 2638.09 kg; mass of hydrogen.

a)

b)

Explanation:

a) According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

b)

According to stoichiometry:

1 mole of gives 7 moles of

Thus  moles of will give = of

Mass of

Thus of can be obtained from kg of propane

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