How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium

Chemical reaction:

2HCI + Na —> 2NaCl +H2

4.85 grams of H2 will be produced

Explanation:

Step 1: Data given

Mass of HCl = 175 grams

Molar mass of HCl = 36.46 g/mol

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

2HCI + Na —> 2NaCl +H2

Step 3: Calculate moles HCl

Moles HCl = mass HCl / molar mass HCl

Moles HCl = 175 grams / 36.46 g/mol

Moles HCl = 4.80 moles

Step 4: Calculate moles H2

For 2 moles HCl we need 1 mol Na to produce 2 moles NaCl and 1 mol H2

For 4.80 moles HCl we'll have 4.80/ 2 = 2.40 moles H2

Step 5: Calculate mass H2

Mass H2 = moles H2 * molar mass H2

Mass H2 = 2.40 moles * 2.02 g/mol

Mass H2 = 4.85 grams

4.85 grams of H2 will be produced

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