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3 years ago

H2 + O2 --> __H2O

Chemistry

1 answer:

nekit [7.7K]3 years ago

6 0

Answer:

42.3moles

Explanation:

The chemical equation given in this question is as follows:

H2 + O2 → H2O

However, this equation is unbalanced, the balanced chemical equation is as follows:

2H2 + O2 → 2H2O

This equation shows that;

2 moles of hydrogen gas (H2) will produce 2 moles of water (H2O)

Hence, If 42.3 moles of water (H2O) are produced, 42.3 × 2/2

= 42.3moles of hydrogen was reacted.

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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:

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

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3 years ago

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3 years ago

__H2 + __O2 --> __H2O

H2 + O2 --> __H2O