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Gnesinka [82]
3 years ago
7

For the reaction 2 nh3 + ch3oh → products, how much ch3oh is needed to react with 93.5 g of nh3? 1. 1.31 mol 2. 46.8 mol 3. 2.75

mol 4. 11.3 mol 5. 3.32 mol 6. 88.1 mol 7. 5.50 mol
Chemistry
2 answers:
aalyn [17]3 years ago
8 0

Answer:

3) 2.75 mol

Explanation:

2 nh3 + ch3oh → products

From the equation above, it can be deduced that 2 moles of NH3 reacts with 1 mole of CH3OH to form the products.

Since the answer is needed in moles,

First we convert g of NH3 to moles. The molar mass of NH3, using the atomic weights on the periodic table, = N (14.0) + 3H (3 x 1.0) = 17.0 g/mole.

93.5 g NH3 x (1 mole NH3 / 17.0 g NH3) = 5.5 moles NH3

The equation tells us that 2 moles of NH3 reacts with 1 mole of CH3OH.

2 moles of NH3 = 1 mole of CH3OH

5.5 moles of NH3 = x mole of CH3OH

x mole = (5.5 moles of NH3 * 1 mole of CH3OH) / 2 moles of NH3

x = 2.75 mol

DENIUS [597]3 years ago
6 0
3.2.75 mol is the answer.
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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
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A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

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M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

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Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

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0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

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