How many hydrogen atoms are in 8.90 mol of ammonium sulfide?

2 answers:

5 0

Assuming that the ammonium sulfide formula is (NH4)2<span>S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles.
If you are asked the number of individual atoms you need to multiply it with Avogadro number like this : </span>71.2 moles * 6.02 * 10^23= 4.28 *10^25

kompoz [17]3 years ago

3 0

Answer:

Explanation:

Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles.

If you are asked the number of individual atoms you need to multiply it with Avogadro number like this : 71.2 moles * 6.02 * 10^23= 4.28 *10^25

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In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)

lyudmila [28]

Answer : The mass of $NaBr$ is, 295.323 grams

Solution :

First we have to calculate the moles of $NaNO_3$ .

$\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles$

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g$

Therefore, the mass mass of $NaBr$ is, 295.323 grams

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