The atmosphere is a mixture of gasses.
Answer:
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Answer:
NaOH is the limiting reactant.
204.9 g of sodium phosphate are formed.
51.94 g of excess reactant will remain.
Explanation:
The reaction that takes place is:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
First we <u>convert the mass of both reactants to moles</u>, using their <em>respective molar masses</em>:
- H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol
- NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol
1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.
--
We <u>calculate the produced moles of Na₃PO₄</u> using the <em>limiting reactant</em>:
- 3.75 mol NaOH *
= 1.25 mol Na₃PO₄
Then we <u>convert moles into grams</u>:
- 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g
--
We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:
- 3.75 mol NaOH *
= 1.25 mol H₃PO₄
We substract that amount from the original amount:
- 1.78 - 1.25 = 0.53 mol H₃PO₄
Finally we <u>convert those remaining moles to grams</u>:
- 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g
Answer:
ΔH = -55.92 kJ
Explanation:
<u>Step 1:</u> Data given
1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water
After mixing the temperature rises to 83 °C
Specific heat of the solution = 4.184 J/g °C
Molar mass of NaOH = 40 G/mol
Molar mass of HBr = 80.9 g/mol
<u>Step 2: </u>The balanced equation
NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)
<u>Step 3:</u> mass of NaOH
Mass = moles * Molar mass
Mass NaOH = 1 * 40 g/mol
Mass NaOH = 40 grams
Step 4: Mass of HBr
Mass HBr = 1 mol * 80.9 g/mol
Mass HBr = 80.9 grams
Step 5: Calculate ΔH
ΔH = m*c*ΔT
ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)
ΔH= 220.9 * 4.184 * 60.5
ΔH= 55916.86 J = 55.92 kJ
Since this is an exothermic reaction, the change in enthalpy is negative.
ΔH = -55.92 kJ