In the chemical reaction nahco3 + ch3cooh → ch3coona + h2o +co2, 83 g of sodium bicarbonate reacts with 70 g of acetic acid. which of these is correct?
a) amount of nahco3 is in excess.
b) amount of ch3cooh is in excess.
c) ch3cooh is the limiting reagent.
d) amount of ch3coona is in excess.

MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g

84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with

84g 60g
83g X
X = 59,29g CH₃COOH

We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.

B)Amount of CH3COOH is in excess.

:•)

6.46×10²² formula units

Explanation:

From the question given above, the following data were obtained:

Mass of MgCl₂ = 10.2 g

Number of formula units =?

From Avogadro's hypothesis,

1 mole of MgCl₂ = 6.02×10²³ formula units

But,

1 mole of MgCl₂ = 24 + (35.5×2) = 24 + 71 = 95 g

Thus, we can say:

95 g of MgCl₂ = 6.02×10²³ formula units

Finally, we shall determine the formula units in 10.2 g MgCl₂. This can be obtained as follow:

95 g of MgCl₂ = 6.02×10²³ formula units

Therefore,

10.2 g of MgCl₂ = (10.2 × 6.02×10²³) / 95

10.2 g of MgCl₂ = 6.46×10²² formula units

Thus, 10.2 g of MgCl₂ contains 6.46×10²² formula units