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Masja [62]
3 years ago
8

25 points , hi! please look at the attachment for the question, I'm having a hard time because all I have to do is solve through

those 2 problems but I'm not sure if I solve the fractions first then multiply that result by 100 or what. I asked my teacher which she helped a little but I don't think she understood where I was lost. If you can help I would really appreciate it. Thank you.
This was the set up question:
4. If 100.0g of nitrogen gas (N2) is reacted with 100.0g of hydrogen gas (H2) to form NH3. What is the limiting and excess reactants?
Hint: Convert grams to moles for each reactant and then convert to moles of NH3. You need your balanced equation from answer 1 to determine the mole relationship between each reactant and the product NH3. Use the periodic table to determine the molar mass of all chemical formulas. Fill in the “?” blanks below to show your work.

and in the screenshot it has everything I'm working with and the conclusions I need to draw from it, I can draw the conclusions just fine on my own but I need help solving.

Chemistry
2 answers:
Finger [1]3 years ago
8 0

Answer:

Explanation:

Did u ever get answer

MAXImum [283]3 years ago
7 0

Answer:

To answer the question, we correctly fill the attached screenshot as follows;

  • 3H₂ + N₂ → 2NH₃
  • The molar mass of H₂ = 2 g/mol

100.0 g \ H_2 \times \dfrac{1 \ mol \ H_2}{2 \ g \ H_2} \times \dfrac{2 \ mol \ NH_3}{3 \ mol \ H_2} = 33.\bar 3 \ mol \ NH_3

The molar mass of N₂ = 28 g/mol

100.0 g \ N_2 \times \dfrac{1 \ mol \ N_2}{28 \ g \ N_2} \times \dfrac{2 \ mol \ NH_3}{1 \ mol \ N_2} \approx 7.143 \ mol \ NH_3

A. Therefore, the excess reactant is hydrogen gas H₂ because it makes the most amount of ammonia, NH₃ (33.\bar 3 moles of NH₃)

B. The limiting reactant in nitrogen, N₂, because it is the reactant that makes the least amount of the ammonia, NH₃ (approximately 7.143 mol NH₃)

C. The theoretical yield of ammonia, is the maximum amount of ammonium that can be produced from the reaction between the 100 g of hydrogen gas, H₂, and 100 g of nitrogen gas, N₂ which is given by the amount of ammonia produced by the limiting reactant which is approximately 7.143 mole of NH₃

Explanation:

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Answer:

Explanation:

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?

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first, balance the equation

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two magnesium atoms react with one diatomic oxygen molecule

there is a 1:1 ratio of magnesium to oxygen atoms

but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms

the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.

6 0
3 years ago
If have a volume of 18 L of a gas at a temperature of 272 K and a pressure of 90 atm, what will be the pressure of the gas if ra
Solnce55 [7]

Answer:

P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)

Explanation:

Given:

P₁ = 90 atm                    P₂ = ?

V₁ = 18 Liters(L)              L₂ = 12 Liters(L)      

=> decrease volume => increase pressure

=> volume ratio that will increase 90 atm is (18L/12L)                                                                  

T₁ = 272 Kelvin(K)          T₂ = 274 Kelvin(K)

=>  increase temperature => increase pressure

=> temperature ratio that will increase 90 atm is (274K/272K)

n₁ = moles = constant    n₂ = n₁ = constant

P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)

By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.

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Explanation:

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6 0
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An order is given to administer 57 g of lactulose syrup, often used for treating complications of liver disease. The suspension
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Answer : The volume given to the patient should be, 85.5 mL

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As we are given that the suspension contains 10g/15mL.

Now we have to determine the volume should be given to the patient.

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So, 57 grams of lactulose syrup needed \frac{57g}{10g}\times 15mL=85.5mL volume of solution

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