All categories

4 years ago

how many grams of calcium oxide are producer from the decomposition of 125 grams of calcium carbonate

Chemistry

1 answer:

Dmitrij [34]4 years ago

7 0

70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.

MM = 100.09 56.08

CaCO3 → CaO + CO2

Mass 125 g

a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)

= 1.249 mol CaCO3

b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)

= 1.249 mol CaO

c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g

You might be interested in

A year has 365 days, and a day has 24 hours. If an hour has 60 minutes and a minute has 60 seconds, how many seconds are there i

Cloud [144]

Answer:

31,536,000 seconds

Explanation:

off google

3 0

3 years ago

Read 2 more answers

How many moles are in 456 g of CaCO3

GrogVix [38]

Answer:

4.556 moles

Explanation:

mole=mass/ molar mass

mole= 456/100.08

mole= 4.556

3 0

3 years ago

Given the electron configuration of an atom: 1s2, 2s2, 2p5 , what is the element?

andrew11 [14]

Question

<em>Given the electron configuration of an atom: 1s2, 2s2, 2p5 , what is the element? </em>

Answer:

<em>A.) Fluorine </em>

Hope this helps!

4 0

3 years ago

I will give thanks, profile thanks, brainliest and 5 star

liubo4ka [24]

Answer: E.) C2N2H8

Explanation:

4 0

3 years ago

Read 2 more answers

MAVERICK [17]

The equilibrium constant (K) : 11.85

<h3>Further explanation</h3>

Given

Reaction

N₂(g) + 3H₂(g) ⇒ 2NH₃(g)

Required

K(equilibrium constant)

Solution

The equilibrium constant (K) is the value of the concentration product in the equilibrium

The equilibrium constant based on concentration (K) in a reaction

pA + qB -----> mC + nD

$\tt K=\dfrac{[C]^m[D]^n}{[A]^p[B]^q}$

For the reaction above :

$\tt K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}\\\\K=\dfrac{0.1^2}{0.25\times 0.15^3}\\\\K=11.85$

6 0

3 years ago