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timurjin [86]
3 years ago
15

Na +H₂O- NaOH +H₂ balance

Chemistry
1 answer:
Whitepunk [10]3 years ago
7 0

Hey there!

Na + H₂O → NaOH + H₂

First, balance O.

One on the left, one on the right. Already balanced.

Next, balance H.

Two on the left, three on the right. Let's add a coefficient of 2 in front of NaOH and a coefficient of 2 in front of H₂O, so we have 4 on each side.

Na + 2H₂O → 2NaOH + H₂

Lastly, balance Na.

One on the left, two on the right. Add a coefficient of 2 in front of Na.

2Na + 2H₂O → 2NaOH + H₂  

This is our final balanced equation.

Hope this helps!

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What are the coefficients to balance the following equation?<br><br> ba+br2=babr2
kaheart [24]

Answer:

Its already balanced

Explanation:

Ba (barium) starts with and ends with 1. Br (bromine) starts with 2 and ends with 2.

6 0
3 years ago
Read 2 more answers
How many moles of carbon are in 3.7 moles of c8h11no2
sleet_krkn [62]

Answer:- 29.6 moles of carbon.

Solution:- We have been given with 3.7 moles of C_8H_11NO_2 and asked to calculate the moles of C.

Looking at the formula of the compound, there are 8 carbons in it means 1 mol of he compound has 8 moles of C. So, if we multiply the given moles of the compound by 8 then we get the moles of C.

3.7molC_8H_11NO_2(\frac{8molC}{1molC_8H_11NO_2})

= 29.6 mol C

Hence. there are 29.6 moles of C in 3.7 moles of  C_8H_11NO_2  .

8 0
3 years ago
Limiting reactants would appreciate the help
Vanyuwa [196]

Answer:

Explanation:

The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.

To identify the limiting reactant ...

- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).

- convert all given reactant values to moles

- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.  

Your Problem:

Given:        3Ba  +  N₂  => Ba₃N₂

               22.6g    4.2g        ?

moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba

moles N₂ =>    4.2g/14.007g/mol= 0.150 mole N₂

Part A: Determining the Limited Reactant

  • Divide each mole value by respective coefficient ... smallest value is Limiting  Reactant.

Barium => 0.165/3 = 0.055  <=> (Limiting Reactant)

Nitrogen => 0.15/1 = 0.15

  • Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.

Part B: Max (theoretical) amount of Ba₃N₂ produced:

<em>Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant.     </em>

                   3Ba        +          N₂         =>     Ba₃N₂    (3:1 rxn ratio for Ba:Ba₃N₂)

moles      0.165mole        0.150mole         1/3(0.165)mole = 0.055mole Ba₃N₂

                                                                    = 0.055mol(440g/mol) Ba₃N₂

                                                                    = 24.2 grams Ba₃N₂ (as based

                                                                     upon Barium as Limiting Reactant)

Part C: Excess N₂ remaining after reaction stops:

From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) =  moles of N₂ used.

moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used  

∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.

mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.

                   

5 0
2 years ago
Calculate the mass percent for all components in a solution containing the following. 0.350 Kg of water, 5.4 moles of ammonia an
Alina [70]

Answer:

  • % Water = 54.96%
  • % Ammonia = 0.14%
  • % Cobalt (II) Nitrate = 30.62%

Explanation:

To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:

  • Mass Water ⇒ 0.350 kg Water = 350 g water
  • Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol =  91.8 g
  • Mass cobalt (II) nitrate ⇒ 195.0 g

Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate

  • Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g

To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:

  • % Water ⇒ 350/636.8 * 100% = 54.96%
  • % Ammonia ⇒ 91.8/636.8 * 100% = 0.14%
  • % Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%
8 0
3 years ago
A 30ml sample of a liquid has a mass of 50 grams. What is the density of the liquid?
mash [69]

Answer:

<h2>Density = 1.67 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

mass = 50 g

volume = 30 mL

Substitute the values into the above formula and solve for the density

That's

Density =  \frac{50}{30}  \\  =  \frac{5}{3}  \\  = 1.66666...

Wr have the final answer as

<h3>Density = 1.67 g/mL</h3>

Hope this helps you

5 0
3 years ago
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