20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of NO.

A.
True

B.
False

The statement is false as 85675 grams of nitrogen dioxide is produced instead of 20 grams.

Explanation:

Balance equation for the reaction:

2NO + ⇒ 2N

Data given:

mass of nitrogen produced = 20 gram

mass of oxygen = 29.8 litres or 29800 grams

0xygen is the limiting reagent so,

number of moles of oxygen: (atomic mass of 1 mole 32 grams/mole

number of moles =

putting the values in the equation:

 =  

  = 931.5 moles

1 mole of oxygen reacts to give 2 moles of NO2

931.5 moles of oxygen will give x moles

=

1862.5 moles of NO2 produced

in grams

mass = number of moles x atomic mass

mass = 1862.5 x 46.00

          = 85675 grams

The statement is false as 85675 grams of nitrogen dioxide is produced instead of 20 grams as said in question.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml