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3 years ago

A 31.3-g sample of ammonium carbonate contains ________ mol of ammonium ions.

Chemistry

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:

$n_{NH_4}=0.652molNH_4$

Explanation:

Hello,

Ammonium carbonate has the following formula:

$(NH_4)_2CO_3$

In this manner, the moles are computed by knowing its molecular mass:

$M_{(NH_4)_2CO_3}=14*2+1*8+12*1+16*3=96g/mol$

Thus, by applying a mass-mole-ions relationship, one obtains:

$ions_{(NH_4)_2CO_3}=31.3g(NH_4)_2CO_3*\frac{1mol(NH_4)_2CO_3}{96g(NH_4)_2CO_3}*\frac{2molNH_4}{1mol(NH_4)_2CO_3}\\n_{NH_4}=0.652molNH_4$

Best regards.

Leona [35]3 years ago

7 0

We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows:

31.3 g (NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol ammonium ions

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The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

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Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

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Explanation:

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Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

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