Question

100 POINTS! Please help! (silly answers will be reported)

How many grams of NO2 can be produced from 15.0 g of NO and 5.60 g of O2 according to the equation:

2 NO + O2
2NO2

Answer 1

According to the balanced chemical equation you provided, 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. The molar mass of NO is 30.01 g/mol and the molar mass of O2 is 32.00 g/mol. This means that 15.0 g of NO is equivalent to 0.5 moles and 5.60 g of O2 is equivalent to 0.175 moles.
Since the ratio of NO to O2 in the reaction is 2:1, the limiting reactant in this case is O2. This means that all of the available O2 will be consumed in the reaction and only 0.35 moles (0.175 x 2) of NO will be consumed.
The molar mass of NO2 is 46.01 g/mol, so 0.35 moles of NO2 has a mass of approximately 16.1 grams.
Approximately 16.1 grams of NO2 can be produced from the given amounts of NO and O2.

Answer 2

16.1 g

Step-by-step explanation:

We want to find how many grams of NO₂ can be produced from 15.0 g of NO and 5.60 g of O₂ according to the balanced chemical equation:

`\sf 2 NO + O_2\;\; \longrightarrow \;\;2NO_2`

First, convert the given masses of both reactants, NO and O₂, to moles using their respective relative formula masses
`\sf(M_r)`.

Relative formula masses:

• 
  `\sf M_r\;of\;NO: 30.01\;g/mol`
• 
  `\sf M_r \; of\; O_2: 31.999\;g/mol`

Therefore:

`\sf moles\;of\;NO=(mass\;(g))/(M_r)=(15.0)/(30.01)=0.500`

`\sf moles\;of\;O_2=(mass\;(g))/(M_r)=(5.60)/(31.999)=0.175`

Now look at the ratio of the reactants:

2 mol NO : 1 mol O₂ = 0.5 mol NO : 0.25 mol O₂

There are only 0.175 moles of O₂ (instead of 0.25 moles), so the O₂ will run out first. It is the limiting reactant.

Use the moles of the limiting reactant to calculate the mass of the product, remembering to use the molar ratio between the limiting reactant and the product.

Limiting reactant : product = 1 mol O₂ : 2 mol NO₂

Therefore, 0.175 mol O₂ will make 0.35 mol NO₂.

Finally, convert the moles of NO₂ to grams:

`\begin{aligned}\sf Mass\;of\; NO_2 &= \sf moles * M_r \\&= \sf 0.35 * 46.0055\\ &= \sf 16.1\; g\end{aligned}`