Taking into account the reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2 PH₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Ca₃P₂:1 mole
- H₂O: 6 moles
- Ca(OH)₂: 3 moles
- PH₃: 2 moles
The molar mass of the compounds is:
- Ca₃P₂: 182 g/mole
- H₂O: 18 g/mole
- Ca(OH)₂: 74 g/mole
- PH₃: 34 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Ca₃P₂: 1 mole ×182 g/mole= 182 grams
- H₂O: 6 moles× 18 g/mole= 108 grams
- Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
- PH₃: 2 moles ×34 g/mole= 68 grams
<h3>Correct statements</h3>
Then, by reaction stoichiometry, you can observe that:
- one mole of Ca₃P₂ produces 2 mol of PH₃.
- the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.
Learn more about the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
<h3><u>Answer;</u></h3>
10.80 ° C
<h3><u>Explanation;</u></h3>
From the information given;
Initial temperature of water = 24.85°C
Final temperature of water = 35.65°C
Mass of water = 1000 g
The specific heat of water ,c = 4.184 J/g °C.
The heat capacity of the calorimeter = 695 J/ °C
Change in temperature ΔT = 35.65°C - 24.85°C
= 10.80°C
Rr - Round eyes. the two alleles are R and r, and if R is dominant, the phenotype will be round
Formula : BaI₂. <span>barium iodide</span>
Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of 
require=
of
Thus is the limiting reagent as it limits the formation of product and acts as the excess reagent. (10.0-0.0787)= 9.92 moles of are left unreacted.
Mass of 
Thus 19.9 g of remains unreacted.
