All categories

2 years ago

How many moles of CF4 can be produced when 8.95 mol C reacts with 7.88 mol F2?

Chemistry

1 answer:

Molodets [167]2 years ago

8 0

The remaining moles of C is 5.01 moles while the remaining moles of F₂ is 0.

<h3>Reaction between Carbon and Fluorine </h3>

The reaction between carbon and Fluorine is given as;

C + 2F₂ -------> CF₄

1 : 2 1

from the reaction above,

2 moles of F₂ requires 1 mole of C

7.88 mole of F₂ will require: 7.88/2 = 3.94 moles of C and 3.94 moles of CF₄.

The remaining moles of C = 8.95 - 3.94 = 5.01 moles while the remaining moles of F₂ is 0.

Learn more about moles here: brainly.com/question/15356425

#SPJ1

You might be interested in

The hormone adrenaline can affect only cells with

-Dominant- [34]

Answer:

appropriately shaped receptors

Explanation:

6 0

3 years ago

What mass of hydrochloric acid (in grams) can 2.7 g of sodium bicarbonate neutralize? (Hint: Begin by writing a balanced equatio

Julli [10]

Answer:

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

Explanation:

Step 1: Data given

Mass of sodium bicarbonate = 2.7 grams

Step 2: The balanced equation

HCl + NaHCO3 ⇔ NaCl + H2O + CO2

Step 3: Calculate moles NaHCO3

moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles

Step 4: Calculate moles HCl

For 1 mol NaHCO3 we need 1 mol HCl

For 0.032 moles NaHCO3 = 0.032 moles HCl

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass HCl

mass HCl = 0.032 * 36.46 g/mol= 1.17 grams

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

3 0

3 years ago

Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g)

Marrrta [24]

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

3(46) kg-------------- 2(63) kg

x --------------- 7600000 kg

x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30

2 NO(g) + O2(g)---2 NO2(g)

2(30) ------------------2(46)

x ---------------- 8323809.5 kg

x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

4(17) -------------------- 4(30)

x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

4 0

3 years ago

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

= 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

7 0

2 years ago

Hi, someone knows how to balance the following equation, and if you can thank you if you can explain how that is done:

kari74 [83]

To begin with, the equation given is not correct.
Correct equation is : CaCO3 + HCl ---> CaCl2 + H2O + CO2
It's CaCl2 not CaCl because Ca has a valency of 2

LHS RHS
CaCO3 + HCl ---> CaCl2 + H2O + CO2
First of all, to balance the equation you must look at the number of atoms on each side of the equation.
we have 2 H on the RHS and 1 H on the LHS. So, we put a 2 on the LHS

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Check for the LHS: 1 Ca, 1 C, 3 O, 2 H & 2 Cl on the LHS
Now check for the RHS: 1 Ca, 2 Cl, 2 H, 1 C & 3 O

Hope it helped!

7 0

3 years ago