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Answer: 1. 2.7 moles of ammonia are formed
2. 12.0 moles of hydrogen are required
3. 2.0 moles of nitrogen are required
Explanation:
The balanced chemical equation is:
According to stoichiometry:
3 moles of hydrogen form = 2 moles of ammonia
Thus 4.0 moles of hydrogen form = of ammonia
According to stoichiometry:
2 moles of ammonia are formed by = 3 moles of hydrogen
Thus 8.0 moles of ammonia are formed by = of hydrogen
According to stoichiometry:
3 moles of hydrogen react with = 1 mole of nitrogen
Thus 6.0 moles of hydrogen react with = of nitrogen
Answer:
1.67 moles
Explanation:
From the balanced equation of reaction:
1 mole of sulfur dioxide, SO2, is required to produce 3 moles of sulfur, S.
<em>If 1 mole SO2 = 3 moles S, then, how many moles of SO2 would be required for 5 moles S?</em>
Moles of SO2 needed = 5 x 1/3
= 5/3 or 1.67 moles
Hence, <u>1.67 moles of SO2 would be required to produce 5.0 moles of S.</u>
Answer:
The total change in enthalpy for the reaction is - 81533.6 J/mol
Explanation:
Given the data in the question;
Reaction;
HCl + NaOH → NaCl + H₂O
Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J
Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
so, 0.0500 moles of H₂O produced
Volume of solution = 50.mL + 50.mL = 100.0 mL
Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g
now ,
Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal
we know that; The specific heat of water(H₂O) is 4.18 J/g°C.
so we substitute
q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28
q_soln = 2842.4 + 1234.28
q_soln = 4076.68 J
Enthalpy change for the neutralization is ΔH
ΔH = -q_soln / mole of water produced
so we substitute
ΔH = -( 4076.68 J ) / 0.0500 mol
ΔH = - 81533.6 J/mol
Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol