For the reaction 2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3 how many grams of sodium chromate, Na2CrO4, are needed to react completely with 4
1 answer:
Answer:
28.12 g Na2CrO4
Explanation:
To solve this question, we have to work with stoichiometry:
Then, if we have 45.5 g of Silver Nitrate:
We know that for each Na2CrO4's mol, we need 2 AgNo3's moles, so, we can calculate the amount of AgNo3's moles as follows:
and, now we can calculate the grams of sodium chromate as:
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Answer:
91383 J
Explanation:
The equation of the reaction can be represented as:
------>
Given that:
The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.
The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.
=
where:
= enthalpy of reaction
= the difference in the heat capacities of the products and the reactants.
∴
=
=
= 91300 J + (0.605 J.K⁻¹)(435-298.15)K
= 91382.79 J
≅ 91383 J
Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :
a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol