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Margarita [4]
3 years ago
5

If 20.0 g of KOH react with 15.0 g of (NH4)2SO4 according to the following reaction, how many liters of NH3 will be produced at

standard temperature and pressure? (The equation is already balanced.)
2KOH + (NH4)2SO4 → K2SO4 + 2NH3 + 2H2O
Chemistry
1 answer:
Serga [27]3 years ago
5 0
Yes equation is ballanced, so you have to apply stoichiometry since you having two masses. find limiting reagent and use mole ratio to find the moles of AMMONIA
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What percent of I-125 has decayed if there are 37.5g of the original sample left?
garri49 [273]

Answer:

[(x-37.5)/x]*100%

62.5%  (assuming the original sample weighs 100.0g)

Explanation:

Let's say that the original sample is x

Mass of I-125 which has decayed: x-37.5

Percentage of decayed mass: [(x-37.5)/x]*100%

Please recheck, for this may not be the correct answer

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A sample of sodium reacts completely with 0.497 kg of chlorine, forming 819 g of sodium chloride. What mass of sodium reacted?
dolphi86 [110]

Answer:

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3 years ago
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A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu
icang [17]

Answer : The partial pressure of N_2 and H_2 is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

p_i=X_i\times p_T

X_i=\frac{n_i}{n_T}

So,

p_i=\frac{n_i}{n_T}\times p_T

where,

p_i = partial pressure of gas

X_i = mole fraction of gas

p_T = total pressure of gas

n_i = moles of gas

n_T = total moles of gas

The balanced decomposition of ammonia reaction will be:

2NH_3\rightarrow N_2+3H_2

Now we have to determine the partial pressure of N_2 and H_2

p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T

Given:

n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg

and,

p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T

Given:

n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg

Thus, the partial pressure of N_2 and H_2 is, 216.5 mmHg and 649.5 mmHg

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