Zn + 2hcl → zncl2 + h2
at conditions of standard temperature and pressure, determine how many liters of hydrogen gas are produced by placing a zinc nail with a mass of 2.2g into an excess of hydrochloric acid.

a) 0.0015 l b) 0.75 l c) 1.33 l d) 6.42 l

0.75L

Explanation:

To get the number of liters of hydrogen gas produced, it is very essential to get the number of moles of hydrogen gas produced.

The number of moles of hydrogen gas produced is obtainable and accessible from

the number of moles of zinc reacted.

Hence, we will get the number of moles of zinc reacted at first. To get the number of moles of zinc reacted, we divide the mass of the zinc by the atomic mass of the zinc metal which is 65.38 amu

The number of moles of the zinc is thus: 2.2/65.38 which equals 0.03365 mole

From the equation of chemical reaction, it can be seen that one mole of zinc metal yielded one mole of hydrogen gas. Hence, it is expected that 0.03365mole of zinc will yield 0.03365mole of hydrogen gas.

Now, we proceed to get the volume of the hydrogen gas produced. We know quite well that at S.T.P a mole of a gas occupies a volume of 22.4L. Hence , 0.03365 mole will occupy a volume of 22.4 * 0.03365 which equals 0.75L

0.75 L of H2 is produced.

Explanation:

From the balanced equation given,we can say that:

1 mole of Zn reacts to give 1 mole of H2.

Since it is carried out at standard temperature and pressure, 1 mole of any gas will occupy 22.4 L.Hence:

1 mole of Zn reacts to give 22.4 L of H2.

Here, mass of Zinc = 2.2 g

Molar mass of Zn = 65 g

Number of moles of Zn =

0.0335 mol of Zn reacts to give = 0.0335×22.4 L of H2

                                                     = 0.75 L

c is the answer because when something is left after something evaporates that shows the chemical change.

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