For this reaction, 11.5 g nitrogen monoxide reacts with 9.91 g oxygen gas. nitrogen monoxide (g) + oxygen (g) nitrogen dioxide (g) What is the maximum mass of nitrogen dioxide that can be formed? g What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?

Answer : The mass of nitrogen dioxide is, 17.6 grams

The formula of limiting reagent is,

The mass of excess reagent remains is, 3.74 grams

Explanation : Given,

Mass of = 11.5 g

Mass of = 9.91 g

Molar mass of = 30 g/mol

Molar mass of = 32 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 2 mole of react with 1 mole of

So, 0.383 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Moles of excess reagent remains = 0.309 - 0.192 = 0.117 mol

Now we have to calculate the mass of excess reagent remains.

Molar mass of = 32 g/mole

Now we have to calculate the moles of

From the reaction, we conclude that

As, 2 mole of react to give 2 mole of

So, 0.383 mole of react to give 0.383 mole of

Now we have to calculate the mass of

Molar mass of = 46 g/mole

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.

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