Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
Answer:
0.03947 atm
Explanation:
The relationship between mmHg and atm is given as;
1 atm = 760 mmHg
x atm = 30 mmHg
Upon solving for x;
x * 760 = 30 * 1
x = 30 / 760 = 0.03947 atm
6.07 grams is the theoretical yield of calcium phosphate (Ca₃(PO₄)₂).
<h3>How we calculate mass from moles?</h3>
Mass of any substance can be calculated by using moles as:
n = W/M, where
W = required mass
M= molar mass
Given chemical reaction is:
3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂
From the stoichiometry it is clear that:
3 moles of Ca(NO₃)₂ = produce 1 mole of Ca₃(PO₄)₂
Given mass of Ca(NO₃)₂ = 96.1g
Mole of Ca(NO₃)₂ = 96.1g/164g/mol = 0.5859moles
So, 0.5859 moles of Ca(NO₃)₂ = produce 0.5859×1/3 = 0.0196 moles of Ca₃(PO₄)₂
Required mass of Ca₃(PO₄)₂ will be calculated by using moles as:
W = 0.0196mole × 310g/mole = 6.07 grams
Hence, 6.07 grams is the theoretical yield of calcium phosphate.
To know more about moles, visit the below link:
brainly.com/question/15373263
