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Sedbober [7]
3 years ago
7

What mass (in g) of NH3 must be dissolved in 475 g of methanol to make a 0.250 m solution?

Chemistry
1 answer:
Tema [17]3 years ago
5 0

Answer:

We need 2.02 grams of NH3

Explanation:

Step 1: Data given

Mass of methanol = 475 grams

Molality of the solution = 0.250 M

Molar mass of NH3 =17.03 g/mol

Molar mass of = methanol = 32.04g/mol

Step 2: Calculate moles NH3

Molality = 0.250 m = mol NH3 / 0.475 kg

moles NH3 = 0.250 molal * 0.475 kg

Moles NH3 = 0.11875 moles

Step 3: Calculate the mass of NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.11875 mol * 17.03 g/mol

Mass NH3 = 2.02 grams

We need 2.02 grams of NH3

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m_{H_2SO_4}=81.7gH_2SO_4

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Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

2Al (s) + 3H_2SO_4 (aq)\rightarrow  Al _2(SO4)_3 (aq) + 3H_2 (g)

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2

Best regards.

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