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I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.

In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ (where X is an unknown atom) are 1.75 moles of oxygen atoms.

a) What is the number of Na atoms in the sample?

I figured the number of moles would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.

b) What is the molar mass of X?

I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.

I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:

$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$

Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.

1) How many grams of gaseous oxygen were formed in the process?

I first balanced the reaction thus:

$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$

I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.

2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?

I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.

3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?

I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$

4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?

I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.

I'd appreciate any comments on my attempt above.

I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.

In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ are 1.75 moles of oxygen atoms.

a) What is the number of Na atoms in the sample?

I figured the number would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.

b) What is the molar mass of X?

I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.

I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:

$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$

Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.

1) How many grams of gaseous oxygen were formed in the process?

I first balanced the reaction thus:

$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$

I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.

2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?

I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.

3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?

I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$

4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?

I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.

I'd appreciate any comments on my attempt above.

I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.

In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ (where X is an unknown atom) are 1.75 moles of oxygen atoms.

a) What is the number of Na atoms in the sample?

I figured the number of moles would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.

b) What is the molar mass of X?

I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.

I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:

$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$

Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.

1) How many grams of gaseous oxygen were formed in the process?

I first balanced the reaction thus:

$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$

I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.

2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?

I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.

3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?

I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$

4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?

I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.

I'd appreciate any comments on my attempt above.

1

# Stoichiometry question

I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.

In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ are 1.75 moles of oxygen atoms.

a) What is the number of Na atoms in the sample?

I figured the number would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.

b) What is the molar mass of X?

I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.

I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:

$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$

Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.

1) How many grams of gaseous oxygen were formed in the process?

I first balanced the reaction thus:

$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$

I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.

2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?

I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.

3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?

I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$

4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?

I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.

I'd appreciate any comments on my attempt above.