Balance the following equations.Empty blanks will be considered an implied "I" __ 14 13. 2N_(2)O_(5)+1H_(2)Oarrow underline ( )HNO_(3) 2_(14) 14. 2CrCl_(2)+4Sn_(2)(SO_(4))_(3)arrow 3CrSO_(4)+2SnCl_(3) 15. 1Ca+1As_(2)arrow 2Ca_(3)As_(2) 16. 3Na_(2)CO_(3)+2AlPO_(4)arrow 2Na_(3)PO_(4)+1Al_(2)(CO_(3))_(3)

13. $2N_{2}O_{5}+1H_{2}O\rightarrow 4HNO_{3}$<br />14. $2CrCl_{2}+4Sn_{2}(SO_{4})_{3}\rightarrow 3CrSO_{4}+2SnCl_{3}$<br />15. $3Ca+2As_{2}\rightarrow 2Ca_{3}As_{2}$<br />16. $3Na_{2}CO_{3}+2AlPO_{4}\rightarrow 2Na_{3}PO_{4}+1Al_{CO_{3})_{3}$<br /><br />Explanation:<br />13. To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, we have 2 nitrogen atoms and 5 oxygen atoms on the left side, and 1 hydrogen atom and 3 oxygen atoms on the right side. To balance the oxygen atoms, we need to multiply the number of $HNO_{3}$ molecules by 4. This gives us 4 hydrogen atoms and 12 oxygen atoms on the right side. To balance the hydrogen atoms, we need to multiply the number of $H_{2}O$ molecules by 1. This gives us 2 hydrogen atoms and 5 oxygen atoms on the left side. Therefore, the balanced equation is $2N_{2}O_{5}+1H_{2}O\rightarrow 4HNO_{3}$.<br /><br />14. In this equation, we have 2 chromium atoms, 4 chlorine atoms, and 12 oxygen atoms on the left side, and 3 chromium atoms, 4 oxygen atoms, and 6 chlorine atoms on the right side. To balance the chromium atoms, we need to multiply the number of $CrSO_{4}$ molecules by 3. This gives us 3 chromium atoms and 12 oxygen atoms on the right side. To balance the chlorine atoms, we need to multiply the number of $SnCl_{3}$ molecules by 2. This gives us 4 chlorine atoms on the right side. Therefore, the balanced equation is $2CrCl_{2}+4Sn_{2}(SO_{4})_{3}\rightarrow 3CrSO_{4}+2SnCl_{3}$.<br /><br />15. In this equation, we have 1 calcium atom and 2 arsenic atoms on the left side, and 2 calcium atoms and 2 arsenic atoms on the right side. To balance the calcium atoms, we need to multiply the number of $Ca$ molecules by 3. This gives us 3 calcium atoms on the left side. To balance the arsenic atoms, we need to multiply the number of $As_{2}$ molecules by 2. This gives us 2 arsenic atoms on the left side. Therefore, the balanced equation is $3Ca+2As_{2}\rightarrow 2Ca_{3}As_{2}$.<br /><br />16. In this equation, we have 3 sodium atoms, 3 carbon atoms, and 2 aluminum atoms on the left side, and 2 sodium atoms, 1 aluminum atom, and 3 carbon atoms on the right side. To balance the sodium atoms, we need to multiply the number of $Na_{3}PO_{4}$ molecules by 2. This gives us 6 sodium atoms on the right side. To balance the aluminum atoms, we need to multiply the number of $Al_{2}(CO_{3})_{3}$ molecules by 1. This gives us 1 aluminum atom on the right side. To balance the carbon atoms, we need to multiply the number of $Na_{2}CO_{3}$ molecules by 3. This gives us 3 carbon atoms on the left side. Therefore, the balanced equation is $3Na_{2}CO_{3}+2AlPO_{4}\rightarrow 2Na_{3}PO_{4}+1Al_{2}(CO_{3})_{3}$.