What ions will form when baking soda is dissolved in water?

1 answer:

4 0

Answer:When you add baking powder to water or milk, the alkali and the acidreact with one another and produce carbon dioxide – the bubbles. Sodium bicarbonate is a weak base which is commonly known as baking soda and used in cooking. It weakly ionizes in water: NaHCO3 + H2O → H2CO3 + (OH-) + (Na+). u need to stop deleteing my answers ughh

Explanation:

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Many metals react with oxygen gas to form the metal oxide. for example, calcium reacts as follows. 2 ca(s) + o2(g) → 2 cao(s) yo

aniked [119]

(a) Assuming the amount of O2(g) is not limiting the reaction, a mass of 4.23g of Ca(s) will produce an equal mass of CaO(s), hence it will produce 4.23g of CaO(s). According to their respective molar masses, we have the following CaO molar mass : Molar mass of Ca + Molar mass of O = 40.1 + 16 = 56.1 g/mol 4.23g of Ca will then produce : 4.23 / 56.1 = 0,07540107 mol of CaO. (b) With the same reasonment as above, and assuming the amount of Ca is not limiting, we have : 2.87g of O2 will produce : 2.87 / 56.1 = 0,051158645 mol of CaO. (c) From (a) and (b) answers, we can conclude that the reactant that produces less mol of CaO is limiting the reaction. Hence following the given masses, O2 is the limiting reactant. (d) Knowing the molar mass of CaO is 56.1 g/mol and knowing that O2 is the limiting reactant, we also know the reaction can produce a maximum of 0,051158645 mol of CaO can be produced. So we can conclude we will produce : 56.1 * 0,051158645 = 2.87g of CaO.

8 0

4 years ago

How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl

kogti [31]

Answer:

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:

Mg: 1 mole
HCl: 2 moles
MgCl₂: 1 mole
H₂: 1 mole

Being:

Mg: 24. 31 g/mole
H: 1 g/mole
Cl: 35.45 g/mole

the molar mass of the compounds participating in the reaction is:

Mg: 24.31 g/mole
HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
H₂: 2*1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

Mg: 1 mole* 24.31 g/mole= 24.31 g
HCl: 2 moles* 36.45 g/mole= 72.9 g
MgCl₂: 1 mole* 95.21 g/mole= 95.21 g
H₂: 1 mole* 2 g/mole= 2 g

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

$moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}$