100ml volume of 0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.
Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,
C1 V1 =C2 V2
where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1) HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1) Caoh2 solution.
While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give in the expression we get the volume of the final concentration.
V1= C2 V2/ C1
= 0.0100m . 150ml /0.0150M
= 100ml
The volume of the hcl solution is 100ml.
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Explanation:
Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.
This means that, for 
moles of 
+ moles of 
= 0.908 mol
Since, 2 moles of = 
= 0.454 mol
As it is known that the ideal gas equation is PV = nRT
And, as the temperature and volume were kept constant, so we can write
= 
= 
= 
= 5.2 atm
Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.
Answer:
That means Cu2O is limiting reagent and C is excess reagent
Explanation:
Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.
To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.
<em>Moles Cu2O -Molar mass: 143.09 g/mol-</em>
114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O
<em>Moles C -Molar mass: 12.01g/mol-</em>
11.1g C * (1mol / 12.01g) = 0.924 moles C
<h3>That means Cu2O is limiting reagent and C is excess reagent</h3>
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