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2 years ago

How many grams of NaOH are produced from 20.0 grams of Na2CO3?

Chemistry

1 answer:

natita [175]2 years ago

4 0

Answer:

Hope this helps!

Explanation:

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams

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stellarik [79]

Answer:

Electric current

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Explanation:

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Now, in the case of water like it says in the question. Water is a mixture of hydrogen and oxygen gas. Thus, when electric current is passed through water which has an Electrolyte, it makes the water to decompose into its compound elements which are hydrogen and oxygen.

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2 years ago

What is the percent composition of N H S O in (NH4)2SO4

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Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

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3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

$\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}$ $\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}$ $\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}$ $\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\ \\ \end{gathered}$

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

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