2.c
idk if it is right
just trying to help
<span>54.8 g of MgI2 can be produced.
To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium
Atomic weight of Iodine = 126.90447
Atomic weight of Magnesium = 24.305
Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394
Now determine how many moles of Iodine and Magnesium you have
moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole
moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole
Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent.
So figure out how many moles of magnesium will be consumed by the iodine
0.393997154 mole / 2 = 0.196998577 mole.
This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2
0.196998577 mole * 278.11394 g/mole = 54.78805 g
Round the result to the correct number of significant figures.
54.78805 g = 54.8 g</span>
Answer:
CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
Explanation:
- Dissolution reactions are reactions that occur when a solute either in gaseous, liquid, or solid form dissolves in a solvent such as water to form a solution.
- In this case we are given Calcium carbonate (CaCO3) which undergoes dissolution according to the equation;
- CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
- Then<em><u> the bicarbonate ion combines with two protons from water to form a weak acid H2CO3. The weak acid is then broken down to form CO2 and H2O since its unstable.</u></em>
Explanation:
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