How many grams of N2 can be produced when 6.50 g of O2
reacts?

3.79 g of N2.

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

4NH3 + 3O2 → 2N2 + 6H2O

Next, we shall determine the mass of O2 that reacted and the mass of N2 produced from the balanced equation.

This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Summary:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Finally, we shall determine the mass of N2 produced by reacting 6.50 g of O2.

This can be obtained as follow:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.50 g of O2 will react to produce = (6.50 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were obtained from the reaction..

it is a good conductor

explanation:

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"