Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
Answer:
4.89 mol
Explanation:
Step 1: Write the balanced equation
This is a double displacement reaction.
3 CaSO₄ + 2 AIP ⇒ Ca₃P₂ + AI₂(SO₄)₃
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of CaSO₄ to AlP is 3:2.
Step 3: Calculate the moles of AlP needed to react with 7.33 moles of CaSO₄
We will use the previously established molar ratio.
7.33 mol CaSO₄ × 2 mol AlP/3 mol CaSO₄ = 4.89 mol AlP
Answer:
1)Filtration to remove sand
2)Evaporation to obtain sugar
Explanation:
Add water to the mixture and mix,sugar will dissolve.
Filter out the sand.
Evaporate the remaining solution to obtain the sugar.
