How many grams of fe is needed to react with 35.0 g of sulfuric acid in the following reaction? 3h2so4 + 2fe -> fe2(so4)3 + 3h2

13.3 g

Explanation:

Let's consider the following reaction.

3 H₂SO₄ + 2 Fe → Fe₂(SO₄)₃ + 3 H₂

The molar mass of H₂SO₄ is 98.1 g/mol. The moles of H₂SO₄ in 35.0 g are:

35.0 g × (1 mol / 98.1 g) = 0.357 mol

The molar ratio of H₂SO₄ to Fe is 3:2. The moles of Fe are:

0.357 mol H₂SO₄ × (2 mol Fe/ 3 mol H₂SO₄) = 0.238 mol

The molar mass of Fe is 55.8 g/mol. The mass of Fe is:

0.238 mol × (55.8 g/mol) = 13.3 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

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.