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Harman [31]
3 years ago
12

Which coefficient before sodium bromide (NaBr) balances this chemical equation?

Chemistry
1 answer:
yuradex [85]3 years ago
8 0

We see that in the left-hand side of the equation, the side of the reactants, that we have 2 moles of Na and bromine is in it's diatomic form.

Therefore, we have 2 moles of each of these elements. When we combine the sodium bromide molecule in the products, we are going to want to keep these molar amounts the same. So we are going to need to put a 2 in front of the sodium bromide in order to correctly balance this equation.

So the coefficient for sodium bromide (NaBr) in this equation is 2.

You might be interested in
For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) --
IRISSAK [1]

Answer:

m_{Zn(OH)_2}=38.4g

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO}  =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O}  =0.386molZn(OH)_2

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g

Because the water limits the yielded amount of zinc hydroxide.

Best regards!

5 0
3 years ago
Please help fast please!!!
Natali [406]

Answer:

welcome to the picture

Explanation:

ok let get it ok finish

5 0
3 years ago
Can some body please help me with this Stoichiometry stuff
andriy [413]

Answer:

See explanations

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversations:

1 mole = 1 formula weight = 6.023 x 10²³ particles = 22.4 liters at STP(0⁰, 1atm).

In this problem you are given the equation Na + H₂O => NaOH + H₂. ‘Reading the equation’ there is 1 mole of Na, 1 mole of water, 1 mole of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation Na + H₂O => NaOH + H₂, one could apply a 2 before the Na but all the following formulas would need be increased by a factor of 2. If one applies ½ to the Na then all the following formulas would need be cut in half also and the reaction stoichiometry would still be valid. The fact that the equation is written with coefficients of 1 is that it is in the smallest whole number ratio of coefficients. This then implies the reaction formula is in ‘standard form’. This also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.  

For your 1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

                         Na                      +            H₂O       => NaOH    +         H₂,

Given:      1.76 x 10²⁴ atoms                excess             ---------          ? grams

Convert atoms Na to moles = 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

=2.922moles Na produces=>2.922moles H₂(because moles Na=moles H₂).

Convert moles to grams =>2.922moles H₂  x  2.000 grams H₂/mole H₂

=5.8443 grams H₂  

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again write equation and assign values to each formula unit in the equation.

                         Na                      +            H₂O        =>           NaOH    +    H₂,

Given:            3.5moles                       excess                      ? grams       ----

Since coefficients of balanced std equation are equal then moles Na equals moles of NaOH, that is, 3.5 moles Na produces => 3.5 moles NaOH

Convert moles NaOH to grams => 3.5 moles NaOH x 40 g NaOH/mole NaOH =  140 grams NaOH    

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

                         Na          +            H₂O                              =>       NaOH    +    H₂

Given:           ? atoms          2.75 x 10²⁵ molecules H₂O    =>     NaOH  + H₂  

Convert to moles =>  2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O = 45.658 moles H₂O =>  45.658 moles Na (equal coefficients)

Convert moles Na to atoms Na  =>   45.658 moles Na x 6.023 x 10²³atoms Na/mole Na = 2.75 x 10²⁵ atoms Na.

Note => Problem 3 could have been solved by inspection b/c coefficients are equal, however, always go through a process that you can justify and defend even if it does take longer. Never assume anything. Depend on what you know, not what you 'think' you know.  

Master the mole concept and you master a lot of chemistry! Good luck.

                             

5 0
3 years ago
When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo
Marta_Voda [28]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

b)

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g

7 0
3 years ago
2CO + O2 --> 2002
olga_2 [115]
8 moles I think I’m not sure
4 0
3 years ago
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