There are 3.0 g of NO and 3.0 g of CO reacting. what is the limiting reactant ?

NO will be the limiting reagent.

Explanation:

The balanced equation is:

2 NO + 2 CO → N₂ + 2 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

NO: 2 molesCO: 2 molesN₂: 1 moleCO₂: 2 moles

Being the molar mass of each compound:

NO: 30 g/moleCO: 28 g/moleN₂: 28 g/moleCO₂: 44 g/mole

Then by stoichiometry the following quantities of mass participate in each reaction:

NO: 2 moles* 30 g/mole= 60 gCO: 2 moles* 28 g/mole= 56 gN₂: 1 mole* 28 g/mole= 28 gCO₂: 2 moles* 44 g/mole= 88 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: If 56 grams of CO react with 60 grams of NO, 3 grams of CO react with how much mass of NO?

mass of NO= 3.21 grams

But 3.21 grams of NO are not available, 3 grams are available. Since you have less moles than you need to react with 3 grams of CO, NO will be the limiting reagent.

74.4 ml of millimeters needed to react with  15 grams of baking soda

Explanation:

Given data

n C8 H8 07= 0.8 M

m NaHCO3=15 g

Solving the problem:

From the equation we get,

3x C6 H8 07 = NaHC03

Finding the molar mass for the baking soda

NaHCO3= 23+1+12+3 x16= 84 g / mol

so,

n NaHCO3= m NaHCO3/ M NaHCO3

And then,

n C6 H8 07 =1/3 × n NaHCO3

From the definition of the molarity we get

V C6 H8 O 7=n C6 H8 07/C6 H8 07

=n NaHCO3/3 × C6 H8 07 =

m NaHCO3/ M NaHCO3/ 3 × C6 H8 07

Calculating the equation we get,

V C6 H8 O 7= 15/84/3 ×0.8=5/84 ×0.8= 0.0744L= 74.4 ml

74.4 ml of millimeters needed to react with  15 grams of baking soda

n C8 H8 07= 0.8 M

m NaHCO3=15 g

Solving the problem:

From the equation we get,

3x C6 H8 07 = NaHC03

Finding the molar mass for the baking soda

Nahco3= 23+1+12+3 x16= 84 g / mol

so,

n NaHCO3= m NaHCO3/ M NaHCO3

And then,

n C6 H8 07 =1/3 × n NaHCO3

From the definition of the molarity we get

V C6 H8 O 7=n C6 H8 07/C6 H8 07

=n NaHCO3/3 × C6 H8 07 =

m NaHCO3/ M NaHCO3/ 3 × C6 H8 07

Calculating the equation we get,

V C6 H8 O 7= 15/84/3 ×0.8=5/84 ×0.8= 0.0744L= 74.4 ml

74.4 ml of millimeters needed to react with  15 grams of baking soda