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3 years ago

If you were given 74.00 grams of Al, how many grains of Al2O3 would be produced?

1 answer:

5 0

We are given the starting amount of Al metal to be used. This will be the starting point of the calculations. We do as follows:

74.00 g ( 1 mol / 26.98 g ) ( 1 mol Al2O3 / 2 mol Al ) ( 101.96 g / 1 mol ) = 139.83 g Al2O3 produced

Hope this answers the question. Have a nice day.

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3 years ago

How do you find the amount of moles is .032 grams of water and whats the answer

masha68 [24]

Answer:

$\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}$

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

2 Hydrogen atoms (subscript of 2)
1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

Hydrogen: 1.0079 g/mol
Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

$0.032 \ g \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}$

Multiply. The grams of H₂O will cancel out.

$0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }$

$\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }$

$0.00177627781 \ mol \ H_2 O$

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

$0.0018 \ mol \ H_2 O$

There are about <u>0.0018 moles in 0.032 grams.</u>

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3 years ago