Answer:
decelerating hope this helps❤️
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
Answer:
see explaination
Explanation:
Molecular equation;
2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)
Total ionic equation; . Includes all ions ;
6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)
Net ionic equation; remove common ions from total ionic;
2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)
Answer:
Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Explanation:
Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)
This is how it starts out.
Left:
Right
So the place to start with this equation is to bring the Cls up to 2
Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
But the Nas are now out of kilter.
Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)
Now the right has a problem. There's only 1 Na
Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)
Check it out. It looks like we are done.
C and A because the product of a matter will always be the same and just google A.
