Zinc will react with hydrochloric acid to produce hydrogen gas. zn(s)+2hcl(aq)⟶zncl2(aq)+h2(g)what is the volume of h2 gas produced if 2.00g of zinc is used with an excess of hydrochloric acid in the reaction above at stp? round to 3 significant figures.

0.690 liters is the volume of hydrogen gas produced if 2.00 grams of zinc is used with an excess of hydrochloric acid.

Explanation:

Mole so zinc =

According to reaction, 1 mole of zinc gives 1 mole of hydrogen gas.

Then 0.03077  mole of zinc will give :

of hydrogen gas

Pressure of hydrogen gas ,P= 1 atm

Temperature  of of hydrogen gas ,T= 273.15 K

Volume of hydrogen gas = V = ?

Moles of hydrogen gas = 0.03077 mol

PV = nRT (Ideal gas equation )

V = 0.690 L

0.690 liters is the volume of hydrogen gas produced if 2.00 grams of zinc is used with an excess of hydrochloric acid.

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