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2 years ago

I NEEDD SOMEBODY HELP

Chemistry

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dybincka [34]2 years ago

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How many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?

salantis [7]

Answer:

85.34g of NH3

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Step 2:

Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.

Therefore, 5.02 moles of NH3 is produced from the reaction.

Step 3:

Conversion of 5.02 moles of NH3 to grams. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Number of mole of NH3 = 5.02 moles

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 5.02 x 17

Mass of NH3 = 85.34g

Therefore, 85.34g of NH3 is produced.

3 0

2 years ago

A 3.4 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.9 g. The two substances react, rel

natima [27]

Answer: 2.7 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$NaHCO_3(aq)+CH_3COOH(aq)\rightarrow CH_3COONa(aq)+H_2O(l)+CO_2(g)$

Given: mass of sodium hydrogen carbonate = 3.4 g

mass of acetic acid = 10.9 g

Mass of reactants = mass of sodium hydrogen carbonate+ mass of acetic acid = 3.4 + 10.9= 14.3 g

Mass of reactants = Mass of products in reaction vessel + mass of carbon dioxide (as it escapes)

Mass of carbon dioxide = 14.3 - 11.6 =2.7 g

Thus the mass of carbon dioxide released during the reaction is 2.7 grams.

7 0

2 years ago

An object weighting 9.6 grams is placed in a graduated cylinder displacing the volume from 10.0mL to 13.2 mL

grigory [225]

The volume of object is 3.2 ml

<h3>Explanation:</h3>

Given:

Mass of the object = M = 9.6 g

Initial volume of liquid: $V_1 = 10.0\ ml$

Final volume of liquid after displacement: $V_2 = 13.2\ ml$

Total volume of the displaced object inside a graduating cylinder will be given as difference between the final volume and initial volume of the expanding object.

$V = V_2-V_1 = 13.2 -1 0.0 = 3.2\ ml$

V = 3.2 ml

3 0

2 years ago