Question

chemical reactions jigsaw
your group will complete one of the chemical reactions below. you will complete the information for the remaining five reactions.

reaction: (with products & balanced!)
__ k + __ n₂ →
reaction type: synthesis / combination
reactant name(s): potassium, nitrogen
n₂ lewis structure:
product name(s): potassium nitride
other notes:

reaction: (with products & balanced!)
__ c₂h₈ + __ o₂ →
reaction type: combustion
reactant name(s): propane, oxygen
co₂ lewis structure:
product name(s): carbon dioxide, water
other notes:

reaction: (with products & balanced!)
__ h₂so₄ + __ sr(oh)₂ →
reaction type: double displacement
reactant name(s): sulfuric acid, strontium hydroxide
lewis structure:
product name(s): strontium sulfate, water
other notes:

Explanation:

Step1: Balance the first reaction (K + N₂ → K₃N)

To balance the reaction between potassium (K) and nitrogen gas (N₂) to form potassium nitride (K₃N), we first look at the nitrogen atoms. There are 2 N atoms on the left (in N₂) and 1 N atom in K₃N. So we put a coefficient of 2 in front of K₃N to balance N: \( \text{K} + \text{N}_2
ightarrow 2\text{K}_3\text{N} \). Now, for K, there are 6 K atoms on the right (2×3), so we put a coefficient of 6 in front of K: \( 6\text{K} + \text{N}_2
ightarrow 2\text{K}_3\text{N} \).

Step2: Balance the second reaction (C₂H₈ + O₂ → CO₂ + H₂O)

First, let's correct the formula: \( \text{C}_2\text{H}_8 \) is actually \( \text{C}_3\text{H}_8 \) (propane). For propane (\( \text{C}_3\text{H}_8 \)) combustion, the products are \( \text{CO}_2 \) and \( \text{H}_2\text{O} \). Balance C: 3 C on left, so 3 \( \text{CO}_2 \) on right: \( \text{C}_3\text{H}_8 + \text{O}_2
ightarrow 3\text{CO}_2 + \text{H}_2\text{O} \). Balance H: 8 H on left, so 4 \( \text{H}_2\text{O} \) (4×2=8) on right: \( \text{C}_3\text{H}_8 + \text{O}_2
ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \). Balance O: Right side has 3×2 + 4×1 = 10 O. So \( \text{O}_2 \) needs a coefficient of 5 (5×2=10): \( \text{C}_3\text{H}_8 + 5\text{O}_2
ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \).

Step3: Balance the third reaction (H₂SO₄ + Sr(OH)₂ → SrSO₄ + H₂O)

This is a double displacement (neutralization) reaction. For \( \text{H}_2\text{SO}_4 \) (sulfuric acid) and \( \text{Sr(OH)}_2 \) (strontium hydroxide) reacting to form \( \text{SrSO}_4 \) (strontium sulfate) and \( \text{H}_2\text{O} \) (water). Balance H and O: On the left, 2 H from \( \text{H}_2\text{SO}_4 \) and 2 H from \( \text{Sr(OH)}_2 \), total 4 H. On the right, each \( \text{H}_2\text{O} \) has 2 H, so we need 2 \( \text{H}_2\text{O} \)? Wait, no: \( \text{H}_2\text{SO}_4 + \text{Sr(OH)}_2
ightarrow \text{SrSO}_4 + 2\text{H}_2\text{O} \). Let's check: Sr: 1 on left, 1 on right. S: 1 on left, 1 on right. O: 4 (from \( \text{H}_2\text{SO}_4 \)) + 2 (from \( \text{Sr(OH)}_2 \)) = 6. On right: 4 (from \( \text{SrSO}_4 \)) + 2 (from 2\( \text{H}_2\text{O} \)) = 6. H: 2 (from \( \text{H}_2\text{SO}_4 \)) + 2 (from \( \text{Sr(OH)}_2 \)) = 4, and 2×2=4 from 2\( \text{H}_2\text{O} \). So balanced: \( \text{H}_2\text{SO}_4 + \text{Sr(OH)}_2
ightarrow \text{SrSO}_4 + 2\text{H}_2\text{O} \), so coefficients are 1, 1, 1, 2.

Answer:

1. \( 6\text{K} + \text{N}_2

ightarrow 2\text{K}_3\text{N} \)

1. \( \text{C}_3\text{H}_8 + 5\text{O}_2

ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \) (corrected formula for propane)

1. \( \text{H}_2\text{SO}_4 + \text{Sr(OH)}_2

ightarrow \text{SrSO}_4 + 2\text{H}_2\text{O} \)