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GuDViN [60]
3 years ago
8

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

Chemistry
1 answer:
solong [7]3 years ago
6 0

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

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Barium sulfate is made by the following reaction.
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Answer:

                      %age Yield =   96 %

Explanation:

                    The balance chemical equation for given double replacement reaction is,

                      Ba(NO₃)₂ + Na₂SO₄ → BaSO₄ + 2 NaNO₃

Step 1: <u>Calculate moles of Ba(NO₃)₂:</u>

Moles  =  Mass / M.Mass

Moles  =  75.1 g / 261.33 g/mol

Moles  =  0.2873 moles of Ba(NO₃)₂

Step 2: <u>Find out moles of BaSO₄ formed:</u>

According to balance chemical equation,

                  1 mole of Ba(NO₃)₂ produced  =  1 mole of BaSO₄

So,

        0.2873 moles of Ba(NO₃)₂ will produce  =  X moles of BaSO₄

Solving for X,

                      X =  0.2873 mol × 1 mol / 1 mol

                       X =  0.2873 moles of BaSO₄

Step 3: Calculate Theoretical Mass of BaSO₄:

Mass  =  Moles × M.Mass

Mass  =  0.2873 mol × 233.38 g/mol

Mass  = 67.07 g of BaSO₄

Step 4: <u>Calculate %age Yield as:</u>

                 Theoretical Yield  =  67.07 g

                  Actual Yield  =  64.4 g

                  %age Yield  =  <u>???</u>

Formula Used:

                   %age Yield  =  (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

                   %age Yield  =  (64.4 g ÷ 67.07 g) × 100

                   %age Yield =  96.01 % ≈ 96 %

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