Answer
Answer 1 : 28.9 g of CO is needed.
Answer 2 : Six moles of [tex]H_{2}O[/tex] over Nine moles of [tex]O_{2}[/tex]
Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of [tex]Fe_{2}O_{3}[/tex].
Answer 4 : Mass of [tex]O_{2}[/tex] = (150 × 3 × 31.998) ÷ (232.29 × 1) grams
Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.
Solution
Solution 1 : Given,
Given mass of [tex]Fe_{2}O_{3}[/tex] = 55 g
Molar mass of [tex]Fe_{2}O_{3}[/tex] = 159.69 g/mole
Molar mass of CO = 28.01 g/mole
Moles of [tex]Fe_{2}O_{3}[/tex] = [tex]\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}[/tex] = [tex]\frac{55 g}{159.69 g/mole}[/tex] = 0.344 moles
Balanced chemical reaction is,
[tex]Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)[/tex]
From the given reaction, we conclude that
1 mole of [tex]Fe_{2}O_{3}[/tex] gives → 3 moles of CO
0.344 moles of [tex]Fe_{2}O_{3}[/tex] gives → 3 × 0.344 moles of CO
= 1.032 moles
Mass of CO = Number of moles of CO × Molar mass of CO
= 1.032 × 28.01
= 28.90 g
Solution 2 : The balanced chemical reaction is,
[tex]2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O[/tex]
From the given reaction, we conclude that the Six moles of [tex]H_{2}O[/tex] over Nine moles of [tex]O_{2}[/tex] is the correct option.
Solution 3 : The balanced chemical reaction is,
[tex]4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}[/tex]
From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of [tex]Fe_{2}O_{3}[/tex].
Solution 4 : Given,
Given mass of [tex]Zn(ClO_{3})_{2}[/tex] = 150 g
Molar mass of [tex]Zn(ClO_{3})_{2}[/tex] = 232.29 g/mole
Molar mass of [tex]O_{2}[/tex] = 31.998 g/mole
Moles of [tex]Zn(ClO_{3})_{2}[/tex] = [tex]\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}[/tex] = [tex](\frac{150\times 1}{232.29})moles[/tex]
The balanced chemical equation is,
[tex]Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}[/tex]
From the given balanced equation, we conclude that
1 mole of [tex]Zn(ClO_{3})_{2}[/tex] gives → 3 moles of [tex]O_{2}[/tex]
[tex](\frac{150\times 1}{232.29})moles[/tex] of [tex]Zn(ClO_{3})_{2}[/tex] gives → [tex][(\frac{150\times 1}{232.29})\times 3] moles[/tex] of [tex]O_{2}[/tex]
Mass of [tex]O_{2}[/tex] = Number of moles of [tex]O_{2}[/tex] × Molar mass of [tex]O_{2}[/tex] = [tex][(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams[/tex]
Therefore, the mass of [tex]O_{2}[/tex] = (150 × 3 × 31.998) ÷ (232.29 × 1) grams
Solution 5 : Given,
Number of moles of [tex]Na_{2}SO_{4}[/tex] = 4.2 moles
Balanced chemical equation is,
[tex]H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}[/tex]
From the given chemical reaction, we conclude that
1 mole of [tex]Na_{2}SO_{4}[/tex] obtained from 2 moles of NaCN
4.2 moles of [tex]Na_{2}SO_{4}[/tex] obtained → 2 × 4.2 moles of NaCN
Therefore,
The moles of NaCN needed = 2 × 4.2 = 8.4 moles
The number of moles is obtained by dividing the mass by the molar mass.
1) The reaction equation is;
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Number of moles of Fe2O3 reacted = 55.0 g/160 g/mol =0.34
From the balanced reaction equation;
1 moles of Fe2O3 reacts with 3 moles of CO
0.34 moles of Fe2O3 reacts with 0.34 moles × 3 moles/1 moles
= 1.02 moles of CO
Mass of CO = 1.02 moles × 28 g/mol = 28.9 g CO
2) Looking at the balanced chemical reaction equation as shown in the question, the valid mole ratio is six moles of H two O over nine moles of O two.
3) The balanced reaction equation is; 4 Fe +3 O2 → 2Fe2O3. From this we can see that the mole ratio to determine the mass of Fe from a known mass of Fe2O3 is four over two.
4) Given that the balanced reaction equation is; Zn(ClO3)2 = ZnCl2 + 3 O2
The equation that could be used to calculate the mass of oxygen formed is;
(150 × 3 × 31.998) ÷ (232.29 × 1) grams
5) The balanced reaction equation is; H2SO4 + 2NaCN = 2HCN + Na2SO4
From the reaction equation;
2 moles of NaCN yields 1 mole of Na2SO4
4.2 moles of NaCN yields 4.2 moles × 1 mole /2 moles
= 2.1 mol NaCN
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