Question

when the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bold type?
sn + naoh → na₂sno₂ + h₂
options: 1, 2, 3, 4, 5
question 2
1 pts
when the following equation is balanced using the smallest possible integers, what will be the coefficient for the substance that has been emboldened?
na₂s₂o₃ + i₂ → nai + na₂s₄o₆
options: 1, 2, 3, 4, 6

Explanation:

Question 1 (Sn + NaOH → Na₂SnO₂ + H₂)

Step1: Balance Sn

Left: 1 Sn; Right: 1 Sn (in Na₂SnO₂). Sn is balanced.

Step2: Balance Na

Left: 1 Na (in NaOH); Right: 2 Na (in Na₂SnO₂). So put 2 in front of NaOH: \( \text{Sn} + 2\text{NaOH}
ightarrow \text{Na}_2\text{SnO}_2 + \text{H}_2 \)

Step3: Balance O

Left: 2 O (from 2 NaOH); Right: 2 O (in Na₂SnO₂). O is balanced.

Step4: Balance H

Left: 2 H (from 2 NaOH); Right: 2 H (in H₂). H is balanced.
Now the equation is \( \text{Sn} + 2\text{NaOH}
ightarrow \text{Na}_2\text{SnO}_2 + \text{H}_2 \). The coefficient of \( \mathbf{H_2} \) is 1? Wait, no, wait: Wait, let's recheck. Wait, original equation: Sn + NaOH → Na₂SnO₂ + H₂. Wait, maybe I made a mistake. Wait, let's do oxidation states. Sn: 0 → +2 (in Na₂SnO₂, Sn is +2: Na is +1, O is -2, so 2(+1) + Sn + 2(-2) = 0 → Sn = +2). H in NaOH: +1 → 0 (in H₂). So Sn is oxidized (loses 2 e⁻), H is reduced (each H⁺ gains 1 e⁻, so 2 H⁺ gain 2 e⁻). So electrons lost = electrons gained. So Sn (loses 2 e⁻) and H₂ (gains 2 e⁻). So the ratio of Sn to H₂ is 1:1? Wait, no, let's balance again. Let's write the equation:

Sn + NaOH → Na₂SnO₂ + H₂

Let’s list atoms:

Sn: 1 (left) vs 1 (right) – balanced.

Na: 1 (left) vs 2 (right) – need 2 NaOH.

O: 1 (left, from NaOH) vs 2 (right, from Na₂SnO₂) – wait, 2 NaOH has 2 O, Na₂SnO₂ has 2 O – balanced.

H: 1 (left, from NaOH) vs 2 (right, from H₂) – wait, 2 NaOH has 2 H, so H on left is 2, right is 2 (in H₂) – balanced. Wait, so after putting 2 in NaOH:

Sn + 2 NaOH → Na₂SnO₂ + H₂

Now check all atoms:

Sn: 1 = 1 ✔️

Na: 2 = 2 ✔️

O: 2 = 2 ✔️

H: 2 = 2 ✔️

So the coefficient of \( \mathbf{H_2} \) is 1? But the options include 1, 2, 3, 4, 5. Wait, maybe the original equation was miswritten? Wait, maybe it's Na₂SnO₃ or another compound? Wait, no, the problem says Na₂SnO₂. Wait, maybe I made a mistake. Wait, let's check again. Wait, NaOH is a base, Sn reacting with base: maybe the correct product is Na₂SnO₃? But the problem says Na₂SnO₂. Anyway, according to the steps above, the coefficient of H₂ is 1? But the options have 1 as an option. Wait, but maybe I messed up. Wait, let's try again.

Alternative approach: Let's assign coefficients. Let Sn be a, NaOH be b, Na₂SnO₂ be c, H₂ be d.

So:

a Sn + b NaOH → c Na₂SnO₂ + d H₂

Sn: a = c

Na: b = 2c

O: b = 2c (since NaOH has 1 O, b NaOH has b O; Na₂SnO₂ has 2 O, c has 2c O)

H: b = 2d (NaOH has 1 H, b has b H; H₂ has 2 H, d has 2d H)

From Sn: a = c. Let's set c = 1, so a = 1. Then Na: b = 2(1) = 2. O: b = 2(1) = 2 ✔️. H: b = 2d → 2 = 2d → d = 1. So the equation is 1 Sn + 2 NaOH → 1 Na₂SnO₂ + 1 H₂. So the coefficient of H₂ is 1. But the options have 1 as the first option. Wait, but maybe the problem was supposed to be Na₂SnO₃? Let's check: If product is Na₂SnO₃, then O: right has 3 O, so b (NaOH) would be 3, then H: 3 = 2d → d = 1.5, not integer. So maybe the problem is correct as Na₂SnO₂. So the coefficient of H₂ is 1? But the options include 1. Wait, but the first option is 1. So maybe that's the answer.

Wait, but maybe I made a mistake. Let's check the oxidation states again. Sn: 0 → +2 (loses 2 e⁻). H in NaOH: +1 → 0 (gains 1 e⁻ per H atom). So for each H₂ molecule, 2 H atoms gain 2 e⁻. So Sn loses 2 e⁻, H₂ gains 2 e⁻. So the ratio of Sn to H₂ is 1:1. So that matches. So the coefficient of H₂ is 1.

But wait, the options for question 1 are 1,2,3,4,5. So the answer for question 1 is 1? But let's check the second question.

Question 2 (Na₂S₂O₃ + I₂ → NaI + Na₂S₄O₆)

Step1: Balance S

Left: 2 S (in Na₂S₂O₃); Right: 4 S (in Na₂S₄O₆). So put 2 in front of Na₂S₂O₃: \( 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2
ightarrow \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \)

Step2: Balance Na

Left: 4 Na (from 2 Na₂S₂O₃); Right: 2 Na (in Na₂S₄O₆) + x Na (in NaI). So 4 = 2 + x → x = 2. So put 2 in front of NaI? Wait, no: Wait, 2 Na₂S₂O₃ has 4 Na. Na₂S₄O₆ has 2 Na, so NaI must have 2 Na. So NaI coefficient is 2? Wait, no: 2 Na₂S₂O₃: 4 Na. Na₂S₄O₆: 2 Na. So NaI: 4 - 2 = 2 Na, so NaI coefficient is 2? Wait, no, NaI has 1 Na per molecule, so coefficient of NaI is 2? Wait, let's do it step by step.

After putting 2 in Na₂S₂O₃:

\( 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2
ightarrow \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \)

Now S: 4 (left) = 4 (right) ✔️.

Na: 4 (left) vs 2 (in Na₂S₄O₆) + y (in NaI) → 4 = 2 + y → y = 2. So NaI coefficient is 2? Wait, no, NaI has 1 Na, so y = 2 → coefficient of NaI is 2.

Now O: 6 (from 2 Na₂S₂O₃: 2*3=6) vs 6 (in Na₂S₄O₆: 6) ✔️.

Now I: 2 (left, from I₂) vs z (in NaI: z*1). So 2 = z → z = 2. So NaI coefficient is 2. Wait, but let's check the equation:

\( 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2
ightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \)

Now check all atoms:

S: 4 (left) = 4 (right) ✔️.

Na: 4 (left) = 2 (NaI) + 2 (Na₂S₄O₆) = 4 ✔️.

O: 6 (left) = 6 (right) ✔️.

I: 2 (left) = 2 (right) ✔️.

Wait, but the emboldened substance is Na₂S₂O₃? Wait, the question says "the substance that has been emboldened" – the emboldened is Na₂S₂O₃? Wait, the original question: "When the following equation is balanced using the smallest possible integers, what will be the coefficient for the substance that has been emboldened? Na₂S₂O₃ + I₂ → NaI + Na₂S₄O₆" – so emboldened is Na₂S₂O₃. So after balancing, the coefficient of Na₂S₂O₃ is 2? Wait, no, wait: Wait, let's re-express the equation.

The reaction is the reaction between thiosulfate and iodine: \( 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2
ightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \). Yes, that's the standard reaction. So the coefficient of \( \mathbf{Na_2S_2O_3} \) is 2. Wait, but the options are 1,2,3,4,6. So 2 is an option.

Wait, but let's confirm:

• S: 2 (left) vs 4 (right) → need 2 Na₂S₂O₃ (2*2=4 S) ✔️.

• Na: 2*2=4 (left) vs 2 (Na₂S₄O₆) + 2 (NaI) = 4 ✔️.

• I: 2 (left) vs 2 (NaI) ✔️.

• O: 2*3=6 (left) vs 6 (Na₂S₄O₆) ✔️.

So the coefficient of \( \mathbf{Na_2S_2O_3} \) is 2.

But wait, the question says "the substance that has been emboldened" – in the equation, Na₂S₂O₃ is emboldened? Wait, the user's image shows: "Na₂S₂O₃ + I₂ → NaI + Na₂S₄O₆" with Na₂S₂O₃ emboldened? Wait, the original question 2: "When the following equation is balanced using the smallest possible integers, what will be the coefficient for the substance that has been emboldened? Na₂S₂O₃ + I₂ → NaI + Na₂S₄O₆" – so the emboldened substance is Na₂S₂O₃. So its coefficient is 2.

But let's go back to question 1. Wait, maybe I made a mistake in question 1. Let's recheck question 1: Sn + NaOH → Na₂SnO₂ + H₂.

Wait, maybe the correct product is Na₂SnO₃? Let's try that. If product is Na₂SnO₃, then:

Sn + NaOH → Na₂SnO₃ + H₂

Balance Sn: 1 = 1.

Na: 1 vs 2 → put 2 NaOH.

O: 1 (NaOH) vs 3 (Na₂SnO₃) → 2 NaOH has 2 O, need 3 O → no, that's not possible. So maybe the original equation is correct as Na₂SnO₂.

Wait, let's check the H again. 2 NaOH has 2 H, which makes 1 H₂ (2 H atoms). So H is balanced. So the equation is balanced as Sn + 2 NaOH → Na₂SnO₂ + H₂. So H₂ has coefficient 1. So question 1 answer is 1, question 2 answer i…

Answer:

(Question 1):
1 (the coefficient of \( \mathbf{H_2} \) is 1)