Answer:
0.767 moles of ammonium phosphate are produced
Explanation:
The reaction of ammona (NH3), with phosphoric acid is:
3 NH3 + H3PO4 → (NH4)3PO4
<em>Where 3 moles of ammonia reacts per mole of H3PO4 to produce 1 mole of ammonium phosphate.</em>
<em />
If 2.3 moles of ammonia reacts, the moles of ammonium phosphate produced if phosphoric acid is in excess are:
2.3 moles NH3 * (1 mole (NH4)3PO4 / 3 moles NH3) =
<h3>0.767 moles of ammonium phosphate are produced</h3>
<em />
Answer:
T₂ = 242 K
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 125 mL
Initial pressure = 1.77 atm
Initial temperature = 213 °C (213 +273 = 486 K)
Final volume = 136 mL
Final pressure = 0.810 atm
Final temperature = ?
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂T₁ / P₁V₁
T₂ = 0.810 atm × 136 mL × 486 K / 1.77 atm× 125 mL
T₂ = 53537.76 atm .mL. K / 221.25 atm . mL
T₂ = 242 K
It is represented by C) KI
K+
I-
So it's KI
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L
