Which of the following is the best name for N,O?

If 16.00 g of O₂ reacts with 80.00 g NO, how many grams of NO₂ are produced? (enter only the value, round to whole number)

Norma-Jean [14]

Answer:

46 g

Explanation:

The balanced equation of the reaction between O and NO is

2 NO + O₂ ⇔ 2 NO₂

Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.

NO: (80 g)/(30.006 g/mol) = 2.666 mol

(2.666 mol)/2 = 1.333

O₂: (16 g)/(31.998 g/mol) = 0.500 mol

(0.500 mol)/1 = 0.500 mol

Since O₂ is smaller, this is the limiting reagent.

The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.

To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.

0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂

1.000 mol × 46.005 g/mol = 46.005 g

You will produce 46 g of NO₂.

6 0

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A 500.0 g block of dry ice (solid CO2, molar mass = 44.0 g) vaporizes at room temperature. Calculate the volume of gas produced

Damm [24]

Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

<h3>Definition of ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Ideal gas law</h3>

An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Volume of gas</h3>

In this case, you know:

P= 1.50 atm
V= ?
n= 500 g× $\frac{1 mole}{44 g}$ = 11.36 moles, being 44 $\frac{g}{mole}$ the molar mass of CO₂
R= 0.082 $\frac{atmL}{molK}$
T= 25 C= 298 K (being 0 C=273 K)