Question

when the following equation is balanced using the smallest possible integers, what will be the coefficient for the substance that has been emboldened?\
ch₃oh + o₂ → co₂ + h₂o\
\
○ 1\
○ 2\
○ 3\
○ 4\
○ 6\
\
question 4\
3 pts\
balance the equation\
mgcl₂ + k₃po₄ → mg₃(po₄)₂ + kcl

Explanation:

First Question (Balancing \( \boldsymbol{\ce{CH3OH + O2 -> CO2 + H2O}} \))

Step1: Balance Carbon

On the left, \( \ce{CH3OH} \) has 1 C. On the right, \( \ce{CO2} \) has 1 C. So C is balanced for now with coefficient 1 for \( \ce{CH3OH} \) and 1 for \( \ce{CO2} \).

Step2: Balance Hydrogen

\( \ce{CH3OH} \) has 4 H (3 from \( \ce{CH3} \), 1 from \( \ce{OH} \)). \( \ce{H2O} \) has 2 H. So to balance H, \( \ce{H2O} \) needs coefficient 2 (since \( 4 \div 2 = 2 \)). Now equation: \( \ce{CH3OH + O2 -> CO2 + 2H2O} \).

Step3: Balance Oxygen

Left: \( \ce{CH3OH} \) has 1 O, \( \ce{O2} \) has \( 2x \) O (x is coefficient of \( \ce{O2} \)). Right: \( \ce{CO2} \) has 2 O, \( \ce{2H2O} \) has 2 O (total 4 O). So left O: \( 1 + 2x = 4 \) → \( 2x = 3 \)? Wait, no—wait, \( \ce{CH3OH} \) coefficient: let's re - evaluate. Let's set \( \ce{CH3OH} \) coefficient as 2. Then C: 2 on left, so \( \ce{CO2} \) coefficient 2. H: \( \ce{CH3OH} \) has 4 H per molecule, 2 molecules have 8 H. So \( \ce{H2O} \) needs coefficient 4 (8 H ÷ 2 H per \( \ce{H2O} \)). Now O: left - \( \ce{2CH3OH} \) has \( 2\times1 = 2 \) O, \( \ce{O2} \) has \( 2x \) O. Right - \( \ce{2CO2} \) has \( 2\times2 = 4 \) O, \( \ce{4H2O} \) has \( 4\times1 = 4 \) O (total 8 O). So left O: \( 2 + 2x = 8 \) → \( 2x = 6 \) → \( x = 3 \). Now balanced equation: \( \ce{2CH3OH + 3O2 -> 2CO2 + 4H2O} \). The emboldened substance is \( \ce{H2O} \), coefficient 4? Wait, no—wait, initial approach. Wait, let's do it properly. Let \( \ce{CH3OH} \) be a, \( \ce{O2} \) be b, \( \ce{CO2} \) be c, \( \ce{H2O} \) be d.
C: \( a = c \)
H: \( 4a = 2d \) → \( d = 2a \)
O: \( a + 2b = 2c + d \). Substitute \( c = a \), \( d = 2a \): \( a + 2b = 2a + 2a \) → \( a + 2b = 4a \) → \( 2b = 3a \). The smallest integer a: let a = 2 (so that 3a is even, since 2b must be integer). Then b = 3, c = 2, d = 4. So \( \ce{H2O} \) (emboldened) has coefficient 4? Wait, but the options have 4. Wait, maybe my first step was wrong. Wait the original equation: \( \ce{CH3OH + O2 -> CO2 + H2O} \). Let's balance step by step:

1. Carbon: 1 in \( \ce{CH3OH} \), 1 in \( \ce{CO2} \) → \( \ce{CH3OH + O2 -> CO2 + H2O} \) (C balanced).
2. Hydrogen: 4 in \( \ce{CH3OH} \), 2 in \( \ce{H2O} \) → need \( \ce{H2O} \) coefficient 2 (4 H). Now: \( \ce{CH3OH + O2 -> CO2 + 2H2O} \).
3. Oxygen: Left: \( \ce{CH3OH} \) (1 O) + \( \ce{O2} \) (2 O). Right: \( \ce{CO2} \) (2 O) + \( \ce{2H2O} \) (2 O) → total 4 O. So left O: 1 + 2x = 4 → 2x = 3 → not integer. So multiply \( \ce{CH3OH} \) by 2: \( \ce{2CH3OH + O2 -> 2CO2 + H2O} \). Now H: 8 in left, so \( \ce{H2O} \) coefficient 4 (8 H). Now: \( \ce{2CH3OH + O2 -> 2CO2 + 4H2O} \). Oxygen: Left: 2 (from \( \ce{2CH3OH} \)) + 2x (from \( \ce{O2} \)). Right: 4 (from \( \ce{2CO2} \)) + 4 (from \( \ce{4H2O} \)) = 8. So 2 + 2x = 8 → 2x = 6 → x = 3. So balanced equation: \( \ce{2CH3OH + 3O2 -> 2CO2 + 4H2O} \). The emboldened \( \ce{H2O} \) has coefficient 4. So the answer for the first question is 4 (option D? Wait the options are 1,2,3,4,6. So 4 is an option.

Second Question (Balancing \( \boldsymbol{\ce{MgCl2 + K3PO4 -> Mg3(PO4)2 + KCl}} \))

Step1: Balance Magnesium

Left: \( \ce{MgCl2} \) has 1 Mg. Right: \( \ce{Mg3(PO4)2} \) has 3 Mg. So \( \ce{MgCl2} \) needs coefficient 3. Now equation: \( \ce{3MgCl2 + K3PO4 -> Mg3(PO4)2 + KCl} \).

Step2: Balance Phosphate (\( \boldsymbol{\ce{PO4^{3 - }}} \))

Left: \( \ce{K3PO4} \) has 1 \( \ce{PO4^{3 - }} \). Right: \( \ce{Mg3(PO4)2} \) has 2 \( \ce{PO4^{3 - }} \). So \( \ce{K3PO4} \) needs coefficient 2. Now equation: \( \ce{3MgCl2 + 2K3PO4 -> Mg3(PO4)2 + KCl} \).

Step3: Balance Potassium (K)

Left: \( \ce{2K3PO4} \) has \( 2\times3 = 6 \) K. Right: \( \ce{KCl} \) has 1 K. So \( \ce{KCl} \) needs coefficient 6. Now equation: \( \ce{3MgCl2 + 2K3PO4 -> Mg3(PO4)2 + 6KCl} \).

Step4: Check Chlorine (Cl)

Left: \( \ce{3MgCl2} \) has \( 3\times2 = 6 \) Cl. Right: \( \ce{6KCl} \) has 6 Cl. Balanced.

Answer:

(First Question): 4 (the option with 4, e.g., if options are numbered as 1:1, 2:2, 3:3, 4:4, 5:6, then 4. Option D: 4)