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Ch.4 - Chemical Quantities & Aqueous Reactions

Chapter 4, Problem 111a

Predict the products and write a balanced molecular equation for each reaction. If no reaction occurs, write 'NO REACTION.' a. HCl(aq) + Hg2(NO3)2(aq) →

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Hi everyone. We're asking for you to get a reaction. And also balance the chemical equation. We have lead nitrate reactor with sodium bromine. So here we have two compounds reacting. So we can switch the ions to form new products. And this is called double displacement reaction. Let's look at each reacting. We have a latin nitrate and it's gonna associate into lead and nitrate. Since you have two nitro on the left side you need to add it to the front of you know three minus. We have studying bromine and that's going to dissociate into any plus and B. R minus. We switched the ions. We're gonna have PB two plus reacting with B. R. Minus. And this will give us PB. We are too solid which is lead bromine. Then we're gonna have an A plus Granted with n. 0. 3 -. And this will give us A. N. 03 which is sodium nitrate. And it's gonna be soluble because nitrates are soluble and now we're gonna have lead nitrate that's sodium bromide and it's gonna give us liberal mind plus sodium nitrate. And so now we need to balance the equation. We have led nitrate roman sodium on both sides. We have one lead on the rack inside to nitrate. One bro. Mean one sodium, we have one letter on the product side, one nitrate mean and one sodium. Since you have one bro meet on the reactive side but two on the product side by two over here you get to And put it to in front of Sunny & Brown. nine. And then we have two sodium over here. This means we need to put it to the front of sodium nitrate so that we can balance this out. And then we have two nitrate over here as well. So we're gonna have led. I trade Plus two sodium bromine and it's gonna your legs remind last two sodium nitrate. Thanks for watching my video and I hope it was helpful.
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