Sodium(Na) is the limiting reagent.
<h3>What is Limiting reagent?</h3>
The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. The precise quantity of reactant required to react with another element may be estimated from the reaction stoichiometry.
How do you identify a limiting reagent?
The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.
Start by writing the balanced chemical equation that describes this reaction
Notice that the reaction consumes 2 moles of sodium metal for every 1 mole of chlorine gas that takes part in the reaction and produces 2 moles of sodium chloride.
now we can see that we have 3 moles of sodium and 3 moles of chlorine, according to question. so, we can say that sodium is the limiting reagent in the given situation.
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Answer:
All are having different valent cation and anion like mono,di and trivalent polyatomic ions .
A. RbNO3
B. K2S
C. NaHS
D. Mg3(PO4)2 formed by divalent Mg+2 and trivalent PO43-
E. CaHPO4
F. PbCO3 , lead is in Pb+2 form
G. SnF2
H. (NH4)2SO4
I. AgClO4
J. BCl3
Answer:
living organisms whose genetic material has been artificially manipulated in a laboratory through genetic engineering
Explanation:
Answer:
Molecular formula is C₂₆H₃₆O₄
Explanation:
The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.
In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H
In 412 g of compound we would have:
(412 . 75.69) / 100 = 311.8 of C
(412 . 15.51) / 100 = 63.9 g of O
(412 . 8.80) / 100 = 36.2 g of H
Now, we can determine the moles of each, that are contained in 1 mol of compound.
312 g / 12 g/mol 26 C
64 g / 16 g/mol = 4 O
36 g / 1 g/mol = 36 H
Molecular formula is C₂₆H₃₆O₄
The molecular weight of this gas will be 45 g/mol .
The state equilibrium equation for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it represents a decent approximation of the activity of many gases under various conditions.
Ideal gas law can be expressed as:
PV =nRT
Calculation of molecular weight by using ideal as law.
Given data:
P = 3 atm
T = 127 °c
Density = 4. 20 grams per liter
PV =nRT
where p is pressure , T is temperature and R is gas constant.
PV = gram / molecular weight RT
Molecular weight = (g/v)( 1/P) RT
Putting the given data in above equation.
Molecular weight =4.20 × 1/ 3× 400 × 0.0831
Molecular weight = 45 g/mol.
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