Answer: 6.162g of Ag2SO4 could be formed
Explanation:
Given;
0.255 moles of AgNO3
0.155 moles of H2SO4
Balanced equation will be given as;
2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)
Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,
Therefore the number of moles of Ag2SO4 produced is given by,
n(Ag2SO4) = 0.255 mol of AgNO3 ×
[0.155mol H2SO4 ÷ 2 mol AgNO3] x
[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]
= 0.0198 mol of Ag2SO4.
mass = no of moles x molar mass
From literature, molar mass of Ag2SO4 = 311.799g/mol.
Thus,
Mass = 0.0198 x 311.799
= 6.162g
Therefore, 6.162g of Ag2SO4 could be formed
Answer:
<h2>A.EARTH PASSES BETWEEN THE SUN AND THE MOON</h2>
Explanation:
When Earth passes directly between Sun and Moon, its shadow creates a lunar eclipse.
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Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
Answer : The
for the reaction is, 54.89 KJ
Solution : Given,
Mass of ZnS = 48.7 g
Molar mass of ZnS = 97.474 g/mole

First we have to calculate the moles of ZnS.

The balanced combustion reaction is,

From the given reaction, we conclude that
As, 2 moles of ZnS gives energy = 220 KJ
So, 0.499 moles of ZnS gives energy = 
Therefore, the
for the reaction is, 54.89 KJ
The value of x in CuSO4. xH2O is 5.
Given that;
Number of moles of anhydrous salt = Number of moles of hydrated salt
Number of moles of anhydrous salt = 319 × 
Mass of hydrated salt = 499 ×
g
Number of moles of anhydrous salt = 319
g/160 g/mol = 0.00199 moles
Number of moles of hydrated salt = 499 ×
g/ 160 + 18x
0.00199 = 499 ×
g/ 160 + 18x
0.00199 (160 + 18x) = 499 ×
0.318 + 0.036x = 0.499
0.036x = 0.499 - 0.318
0.036x = 0.181
x = 0.181/0.036
x = 5
Hence, x in CuSO4. xH2O is 5.
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