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joja [24]
2 years ago
10

PLEASE HELP!!!

Chemistry
1 answer:
trasher [3.6K]2 years ago
4 0

Answer:

(a) oxygen

(b) 154g (to 3sf)

(c) 79.9% (to 3sf)

Explanation:

mass (g) = moles × Mr/Ar

note: eqn means chemical equation

(a)

moles of P = 84.1 ÷ 30.973 = 2.7152 moles

moles of O2 = 85÷2(16) = 2.65625 moles

Assuming all the moles of P is used up,

moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)

moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)

therefore there is insufficient moles of O2 and the limiting reactant is oxygen.

(b)

moles of P2O5 produced

= 2/5 (according to eqn) × 2.7152

= 1.08608moles

mass of P2O5 produced

= 1.08608 × [ 2(30.973) + 5(16) ]

= 154.164g

= approx. 154g to 3 sig. fig.

(c)

% yield = actual/theoretical yield × 100%

= 123/154 × 100%

= 79.870%

= approx. 79.9% (to 3sf)

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Answer : The equilibrium will shift in the left direction.

Explanation :

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The given reaction is:

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As per question, when we are adding N_2  then the concentration of N_2 is increased on product side then the equilibrium will shift in the direction where decrease of concentration of N_2 takes place. Therefore, the equilibrium will shift in the left direction.

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<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                   \displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

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