Hydrogen is a possible future fuel. However, elemental hydrogen is rare, so it must be obtained from a hydrogen- containing compound such as water. If hydrogen were obtained from water, how much hydrogen, in grams, could be obtained from 1.0 L of water? (density of water = 1.0 g/cm)

1.1 × 10² g

Explanation:

First, we will convert 1.0 L to cubic centimeters.

1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³

The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:

1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g

1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:

1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g

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