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3 years ago

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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

1 answer:

fiasKO [112]3 years ago

7 0

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

Explanation:

Balance equation for the reaction:

2NO + $O_{2}$ ⇒ 2N $O_{2}$

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 = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the equation:

= $\frac{29800}{32}$

= 931.5 moles

1 mole of oxygen reacts to give 2 moles of NO2

931.5 moles of oxygen will give x moles

$\frac{2}{1}$ = $\frac{x}{931.5}$

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.

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

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