The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
Oxygen and Hydrogen would most likely form a covalent bond that is polar, or a polar covalent bond. Due to the electronegativity difference between the 2 elements, unequal sharing of the valence electrons will occur, electrons being in closer proximity to Oxygen and farther away from Hydrogen. Resulting in the characteristic partial positive and negative charges to appear for the respective elements.