Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]
Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as
1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.
n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-
Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess
n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.
Total volume= V acid + V base= 28 ml + 60 ml = 98 ml
Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M
The answer is 0.009 M.
Answer:
copper
Explanation:
so for this you can work out the mass for both and compare
so mass = moles × mr
so mass of sodium = 1 × 23= 23 g
and mass of copper = 1 × 63.5= 63.5 g
so copper have more mass :)
To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:
2HCl + Na2CO3 = 2NaCl + H2O + CO2
From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:
0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2
Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.
Answer:
a. Oxygen gas is limiting
Explanation:
hydrogen gas and oxygen gas are reacted to form water
2H₂ + O₂ → 2H₂O
the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂
Given equal masses of H₂ and O₂
assuming 'x' gm for each, no. of moles of each gas =
no. of moles of H₂ = x/2 = 0.5x moles
no.of moles of O₂ = x/32 = 0.031x moles
This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.
