Question

Calculate the mass loss in the following:
${}_1^{2}H{+}_1^{3}H⟶$ ${}_2^{4}He{+}_0^{1}n$
Given the masses: ${}_1^{2}H=2.014$ $amu$, ${}_1^{3}H=3.016$ $amu$; ${}_2^{4}He=4.004$ $amu$, ${}_0^{1}n=1.008$ $amu$.

• A. $0.018$ $amu$
• B. $0.18$ $amu$
• C. $0.0018$ $amu$
• D. $1.8$ $amu$
• E. $18$ $amu$

Answer 1

The correct option is A $0.018$ $amu$

Mass loss = (Mass of reactants) - (Mass of products)

$\Rightarrow \left[\right(2.014)+(3.016\left)\right]-[4.004+1.008]$

$\Rightarrow 5.030-5.0012$

$\Rightarrow$0.018 amu.

$\therefore$Mass loss in the given equation is 0.018 amu.