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klasskru [66]
3 years ago
14

0.2 moles of aluminum has a mass of 5.4 g. what is The molar mass of aluminum​

Chemistry
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

Molar mass = 27 g/mol

Explanation:

Given data;

Number of moles of Al = 0.2 mol

Mass of Al = 5.4 g

Molar mass of Al = ?

Solution:

Formula:

Number of moles = mass/molar mass

By putting values,

0.2 mol = 5.4 g/ molar mass

Molar mass = 5.4 g/ 0.2 mol

Molar mass = 27 g/mol

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(1.8 x 10-2) ÷ (9 x 102) =
8_murik_8 [283]
0.01742919 is the answer because i worked it out
3 0
3 years ago
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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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Answer:

Concentration of chloride ions = 0.584M

Explanation:

The step by step calculations is shown as attached below.

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